Proof Using a Countable Union of Nonempty Open Sets to Prove that $\mathbb{Q}$ Is Infinite I was playing around with an exercise which shows that if $A \neq \varnothing \subseteq \mathbb{R}$ is an open set, then $A \cap \mathbb{Q} \neq \varnothing$. I then extended the problem to a countable union of nonempty open sets, and then intuitively saw that we can always find at least one distinct rational number for each of the sets, and then reasoned that there are infinitely many rational numbers. Naturally, tried to prove it, so here is my attempt.
If $\mathscr{A} = \bigcup_{k=1}^{\infty} A_k \subseteq \mathbb{R}$ is a countable union of nonempty open sets, allowing $x \in \mathscr{A}$ means that there exists a $k \in \mathbb{N}$ such that $x \in A_k$. Since $A_k$ is open, there exists a neighborhood $N_{\epsilon}(x)$ such that $(x-\epsilon,x+\epsilon) \subseteq A_k$. Then between $x-\epsilon$ and $x+\epsilon$, there exists a rational number $q$ between them, so that $q \in N_{\epsilon}(x)$, and therefore $q \in \mathscr{A}$. 
Since we can always find distinct rational numbers for each set in $\mathscr{A}$, and there are infinitely many such sets, we will find infinitely many rational numbers, so $\mathbb{Q}$ is infinite.
Any thoughts or concerns? Thanks in advance.
 A: This proof doesn't work; you need the additional hypothesis that the $A_k$ are pairwise disjoint. Otherwise, you can't know that you're not just picking the same $q$ for each set. For instance, if you had 
$$\mathscr A =\bigcup_{k=1}^{\infty}(0,1)$$
you would find that $\frac{1}2$ is a rational number everywhere in it.
Also, you barely use the fact that $\mathscr A$ is a countable union of other things in your proof. More or less, you find that it is open because it is a union of open sets, and then the rest of the proof proceeds using only that $\mathscr A$ is open.
What does work is proving that there are infinitely many rational numbers in a set like $$\mathscr A = \bigcup_{k=-\infty}^{\infty}(k,k+1)$$
since this is a countable union of disjoint sets, each of which contains a rational number. You can also note that, if $A$ is a non-empty open set and $q_1,q_2,\ldots,q_n$ is a finite sequence of rationals in it, then $A\setminus \{q_1,q_2,\ldots,q_n\}$ is a non-empty open set and thus contains a rational distinct from $q_1,q_2,\ldots,q_n$. One can therefore find that there are infinitely many rationals in every open set.
Of course, if we're really just after proving that the rational numbers are infinite, it would seem easier to note that the integers are infinite and are a subset of the rationals.
