Can the rational numbers be arranged in a sequence? If so, consider any such sequence of all the rational numbers. Show that every real number is a subsequential limit of this sequence.
Since rational number is countably infinite, I see that rational numbers can be arranged in a sequence. But I'm lost how to proceed
Let $q_1, q_2, \ldots$ be an enumeration of the rational numbers.
Consider some real number $r$.
Suppose $r$ has an infinite decimal expansion. By considering all possible truncations of the decimal expansion, you obtain a sequence of rational numbers $q'_1, q'_2 ,\ldots$ that converges to $r$. You can use this to get a subsequence of your original sequence $(q_i)$ that converges to $r$. (Start your new sequence with $q'_1$, which must appear in the original sequence somewhere. If $q'_2$ appears before $q'_1$ in the original sequence, skip it; else append $q'_2$ to your sequence. Do the same for $q'_3$, and so on.)
Suppose $r$ has a finite decimal expansion. You can do a silly trick to give it an infinite decimal expansion (e.g. $1.2 = 1.1999\cdots$, or $-2.348 = -2.3479999\cdots$) and then perform the above procedure.
Hint for second part:
For a real $r$ and an $\epsilon \gt 0$, is $Q \cap (r-\epsilon, r + \epsilon)$ finite?
Let $(q_n) = q_1, q_2, \ldots$ be an enumeration of the rational numbers.
Consider some real number $a$. We can recursively define a strictly increasing function
$\tag 1 \alpha: \Bbb N \to \Bbb N$
as follows:
$\quad \alpha(1) = \text{the smallest natural number } k \text{ such that } \big [ q_k \lt a \big ] \land \big [ a - q_k \lt 1 \big ]$
Assume $\alpha$ is defined on $\{1,2,\dots,n\}$. Define
$\quad \alpha(n+1) = \text{the smallest natural number } k \text{ such that }$
$\quad \quad \big [k \gt \alpha(n) \big ] \land \big [q_k \gt q_{\alpha(n)} \big ] \land \big [q_k \lt a \big ]\land \big [a - q_k \lt \frac{1}{n+1}\big ]$
It is easy to see that the subsequence ${\big ( q_{\alpha(n)}\big )}_{ n \in \Bbb N}$ is strictly increasing and converges to $a$.
