How there can be different kinds of infinities?

This is very simple to see. This is because of:

Claim: A given set $X$ and its power set $P(X)$ can never be in bijection.

Proof: By contradiction. Let $f$ be any function from $X$ to $P(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $P(X)$ i.e., some subset of $S$, is not in the image of $f$. Consider the set:

$T=\{ x\in X: x\not\in f(x) \}.$

For every $x$ in $X$  , either $x$ is in $T$ or not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so the set $T$ can not be the set $f(x)$ (because $x\in T$ but $x\not\in f(x)$). On the other hand, if $x$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again the set $T$ can not be the set $f(x)$. We just proved that $T$ is NOT $f(x)$ for any $x$, and so $f$ is not surjective. Q.E.D.

Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality(Here I am skipping a little; but a use of the Schroeder-Bernstein theorem will fix things).

How there can be different kinds of infinities?

This is very simple to see. This is because of:

Claim: A given set $X$ and its power set $\mathcal{P}(X)$ can never be in bijection.

Proof: By contradiction. Let $f$ be any function from $X$ to $\mathcal{P}(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $\mathcal{P}(X)$, i.e. some subset of $X$, is not in the image of $f$. Consider the set:$$T=\{x\in X:x\not \in f(x)\}.$$For every $x$ in $X$, either $x$ is in $T$ or it is not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so the set $T$ can not be the set $f(x)$ (because $x\in T$ but $x\not \in f(x)$). On the other hand, if $x$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again the set $T$ can not be the set $f(x)$. We just proved that $T$ is NOT $f(x)$ for any $x$, and so $f$ is not surjective. $\mathsf{QED}$

Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality  (here I am skipping a little; but a use of the Cantor-Schröder-Bernstein theorem will fix things).

How there can be different kinds of infinities?

This is very simple to see. This is because of:

Claim: A given set $X$ and its power set $P(X)$ can never be in bijection.

Proof: By contradiction. Let $f$ be any function from $X$ to $P(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $P(X)$ i.e., some subset of $S$, is not in the image of $f$. Consider the set:

$T=\{ x\in X: x\not\in f(x) \}.$

For every $x$ in $X$ , either $x$ is in $T$ or not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so $T$ is not equal to $f(x)$. On the other hand, if $s$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again $T$ is not equal to $f(x)$. Q.E.D.

Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality(Here I am skipping a little; but a use of the Schroeder-Bernstein theorem will fix things).

How there can be different kinds of infinities?

This is very simple to see. This is because of:

Claim: A given set $X$ and its power set $P(X)$ can never be in bijection.

Proof: By contradiction. Let $f$ be any function from $X$ to $P(X)$. It suffices to prove $f$ cannot be surjective. That means that some member of $P(X)$ i.e., some subset of $S$, is not in the image of $f$. Consider the set:

$T=\{ x\in X: x\not\in f(x) \}.$

For every $x$ in $X$ , either $x$ is in $T$ or not. If $x$ is in $T$, then by definition of $T$, $x$ is not in $f(x)$, so the set $T$ can not be the set $f(x)$ (because $x\in T$ but $x\not\in f(x)$). On the other hand, if $x$ is not in $T$, then by definition of $T$, $x$ is in $f(x)$, so again the set $T$ can not be the set $f(x)$. We just proved that $T$ is NOT $f(x)$ for any $x$, and so $f$ is not surjective. Q.E.D.

Thus take any infinite set you like. Then take its power set, its power set, and so on. You get an infinite sequence of sets of increasing cardinality(Here I am skipping a little; but a use of the Schroeder-Bernstein theorem will fix things).