If I understand your question correctly, you are asking:

1. You define entropy as $S=\int\frac{\delta Q}{T}$.
2. Clearly, $T$ is an intensive quantity, as is $\frac{1}{T}$.
3. If $\delta Q$ is extensive, then so is $\frac{\delta Q}{T}$, since a product of an intensive and an extensive quantity is extensive.
4. So, if $\delta Q$ is extensive, then $\int\frac{\delta Q}{T}$, being a sum of extensive quantities, is extensive.
5. But you don't see why $\delta Q$ should be extensive. After all, when we put a hot block of metal into a cold place, the heat is emitted from the surface; there is at least a speed-of-light delay until the interior "knows" that the exterior is being chilled.

I think this is somewhat definitional. If you have a slab of metal, one side of which is cold and the other is hot, then either:

• Your system is not in (internal) thermodynamic equilibrium, so that entropy is not defined, or
• You really mean you have two adjacent slabs of metal, one cold and one hot (but otherwise indistinguishable, so they we mistook them for a single slab).

But then we expect two slabs at different temperatures to have different thermodynamic states. So an extensive quantity will differ between the two of them.

More generally,

1. Suppose a system $S$ is composed of identical components. (These components can have distinguishable subcomponents; thus we might describe a sulfuric acid solution as a collection of drops of fluid (the components), each droplet containing some water and some oleum (the subcomponents).) $S$ will undergo some thermodynamic process.
2. By conservation of energy, $$\delta Q_S=\sum_{s\in S}{\delta Q_s}\tag{1}$$ (This is also somewhat definitional: suppose component $s$ gives off some heat, but then that heat got turned into work "as it travels" to the outside, in component $t$. Then we'd call it "$t$ doing work" instead.)
3. $S$ is in thermodynamic equilibrium iff each component of $S$ is indistinguishable. (This is the definition of thermodynamic equilibrium for composite systems.)
4. In particular, all $s$ start out and end indistinguishable, since $S$ begins and ends in equilibrium.
5. Thus the internal energy at the start and at the end are both independent of $s$.
6. Subtracting, $dU_s$ is independent of $s$.
7. Likewise, if components performed different amounts $\delta W_s$ of work on the exterior, then we could "soup up" our process so that it separates them. But that clearly won't leave the separated components indistinguishable!
8. So $dU_s$ and $\delta W_s$ are independent of $s$.
9. Subtracting, $\delta Q_s$ is independent of $s$ (by the First Law).
10. Substituting into (1) and picking any fixed $s_0$ for the common value, $$\delta Q_S=(\delta Q_{s_0})(\# S)$$ which is clearly extensive.

If I understand your question correctly, you are asking:

1. You define entropy as $S=\int\frac{\delta Q}{T}$.
2. Clearly, $T$ is an intensive quantity, as is $\frac{1}{T}$.
3. If $\delta Q$ is extensive, then so is $\frac{\delta Q}{T}$, since a product of an intensive and an extensive quantity is extensive.
4. So, if $\delta Q$ is extensive, then $\int\frac{\delta Q}{T}$, being a sum of extensive quantities, is extensive.
5. But you don't see why $\delta Q$ should be extensive. After all, when we put a hot block of metal into a cold place, the heat is emitted from the surface; there is at least a speed-of-light delay until the interior "knows" that the exterior is being chilled.

I think this is somewhat definitional. If you have a slab of metal, one side of which is cold and the other is hot, then either:

• Your system is not in (internal) thermodynamic equilibrium, so that entropy is not defined, or
• You really mean you have two adjacent slabs of metal, one cold and one hot (but otherwise indistinguishable, so they we mistook them for a single slab).

But then we expect two slabs at different temperatures to have different thermodynamic states. So an extensive quantity will differ between the two of them.

More generally,

1. Suppose a system $S$ is composed of identical components. (These components can have distinguishable subcomponents; thus we might describe a sulfuric acid solution as a collection of drops of fluid (the components), each droplet containing some water and some oleum (the subcomponents).) These are not statistical-mechanical microstates; they must be large enough for the thermodynamic limit to apply.
2. $S$ will undergo some thermodynamic process.
3. By conservation of energy, $$\delta Q_S=\sum_{s\in S}{\delta Q_s}\tag{1}$$ (This is also somewhat definitional: suppose component $s$ gives off some heat, but then that heat got turned into work "as it travels" to the outside, in component $t$. Then we'd call it "$t$ doing work" instead.)
4. $S$ is in thermodynamic equilibrium iff each component of $S$ is indistinguishable. (This is the definition of thermodynamic equilibrium for composite systems.)
5. In particular, all $s$ start out and end indistinguishable, since $S$ begins and ends in equilibrium. (This is the definition of a thermodynamic process.)
6. Thus the internal energy at the start and at the end are both independent of $s$.
7. Subtracting, $dU_s$ is independent of $s$.
8. Likewise, if components performed different amounts $\delta W_s$ of work on the exterior, then we could "soup up" our process so that it separates them. But that clearly won't leave the separated components indistinguishable!
9. So $dU_s$ and $\delta W_s$ are independent of $s$.
10. Subtracting, $\delta Q_s$ is independent of $s$ (by the First Law).
11. Substituting into (1) and picking any fixed $s_0$ for the common value, $$\delta Q_S=(\delta Q_{s_0})(\# S)$$ which is clearly extensive.

If I understand your question correctly, you are asking:

1. You define entropy as $S=\int\frac{\delta Q}{T}$.
2. Clearly, $T$ is an intensive quantity, as is $\frac{1}{T}$.
3. If $\delta Q$ is extensive, then so is $\frac{\delta Q}{T}$, since a product of an intensive and an extensive quantity is extensive.
4. So, if $\delta Q$ is extensive, then $\int\frac{\delta Q}{T}$, being a sum of extensive quantities, is extensive.
5. But you don't see why $\delta Q$ should be extensive. After all, when we put a hot block of metal into a cold place, the heat is emitted from the surface; there is at least a speed-of-light delay until the interior "knows" that the exterior is being chilled.

I think this is somewhat definitional. If you have a slab of metal, one side of which is cold and the other is hot, then either:

• Your system is not in (internal) thermodynamic equilibrium, so that entropy is not defined, or
• You really mean you have two adjacent slabs of metal, one cold and one hot (but otherwise indistinguishable, so they we mistook them for a single slab).

But then we expect two slabs at different temperatures to have different thermodynamic states. So an extensive quantity will differ between the two of them.

More generally,

1. Suppose a system $S$ is composed of indistinguishable components. (These components can have distinguishable subcomponents; thus we might describe a sulfuric acid solution as a collection of drops of fluid (the components), each droplet containing some water and some oleum (the subcomponents).)
2. By conservation of energy, $$\delta Q_S=\sum_{s\in S}{\delta Q_s}\tag{1}$$
3. $S$ is in thermodynamic equilibrium iff each component of $S$ is indistinguishable. (This is the definition of thermodynamic equilibrium for composite systems.)
4. In particular, all $s$ start out and end indistinguishable, since $S$ begins and ends in equilibrium.
5. Thus the internal energy at the start and at the end are both independent of $s$.
6. Subtracting, $dU_s$ is independent of $s$.
7. Likewise, if components performed different amounts $\delta W_s$ of work on the exterior, then we could "soup up" our process so that it separates them. But that clearly won't leave the separated components indistinguishable!
8. So $dU_s$ and $\delta W_s$ are independent of $s$.
9. Subtracting, $\delta Q_s$ is independent of $s$.
10. Substituting into (1) and picking any fixed $s_0$ for the common value, $$\delta Q_S=(\delta Q_{s_0})(\# S)$$ which is clearly extensive.

If I understand your question correctly, you are asking:

1. You define entropy as $S=\int\frac{\delta Q}{T}$.
2. Clearly, $T$ is an intensive quantity, as is $\frac{1}{T}$.
3. If $\delta Q$ is extensive, then so is $\frac{\delta Q}{T}$, since a product of an intensive and an extensive quantity is extensive.
4. So, if $\delta Q$ is extensive, then $\int\frac{\delta Q}{T}$, being a sum of extensive quantities, is extensive.
5. But you don't see why $\delta Q$ should be extensive. After all, when we put a hot block of metal into a cold place, the heat is emitted from the surface; there is at least a speed-of-light delay until the interior "knows" that the exterior is being chilled.

I think this is somewhat definitional. If you have a slab of metal, one side of which is cold and the other is hot, then either:

• Your system is not in (internal) thermodynamic equilibrium, so that entropy is not defined, or
• You really mean you have two adjacent slabs of metal, one cold and one hot (but otherwise indistinguishable, so they we mistook them for a single slab).

But then we expect two slabs at different temperatures to have different thermodynamic states. So an extensive quantity will differ between the two of them.

More generally,

1. Suppose a system $S$ is composed of identical components. (These components can have distinguishable subcomponents; thus we might describe a sulfuric acid solution as a collection of drops of fluid (the components), each droplet containing some water and some oleum (the subcomponents).) $S$ will undergo some thermodynamic process.
2. By conservation of energy, $$\delta Q_S=\sum_{s\in S}{\delta Q_s}\tag{1}$$ (This is also somewhat definitional: suppose component $s$ gives off some heat, but then that heat got turned into work "as it travels" to the outside, in component $t$. Then we'd call it "$t$ doing work" instead.)
3. $S$ is in thermodynamic equilibrium iff each component of $S$ is indistinguishable. (This is the definition of thermodynamic equilibrium for composite systems.)
4. In particular, all $s$ start out and end indistinguishable, since $S$ begins and ends in equilibrium.
5. Thus the internal energy at the start and at the end are both independent of $s$.
6. Subtracting, $dU_s$ is independent of $s$.
7. Likewise, if components performed different amounts $\delta W_s$ of work on the exterior, then we could "soup up" our process so that it separates them. But that clearly won't leave the separated components indistinguishable!
8. So $dU_s$ and $\delta W_s$ are independent of $s$.
9. Subtracting, $\delta Q_s$ is independent of $s$ (by the First Law).
10. Substituting into (1) and picking any fixed $s_0$ for the common value, $$\delta Q_S=(\delta Q_{s_0})(\# S)$$ which is clearly extensive.