How to show there exists $E$ such that $E \cap K_n$ is dense for every $n$? Let $\Omega$ be a region (nonempty connected open subset of the complex plane).
Let $K_n$ be a sequence of compact sets whose union is $\Omega$, such that $K_n \subset \mathring{K_{n+1}}$ (the interior of $K_{n+1}$)
Now the book says: "Choose a countable set $E \subset \Omega$ such that $E \cap K_n$ is dense in $K_n$ for all $n$"
It's not clear to me why such $E$ exists. 
I know that $K_n$ are compact hence separable, so it exists $E_n$ countable and dense in each $K_n$. So of course $E = \bigcup E_n$ is such that $E \cap K_n$ is dense in every $K_n$, but we need the axiom of choice to show that $E$ is countable, don't we? 
Is there a more immediate way to show that such $E$ exists? Maybe I am overlooking something obvious
Thanks! 
 A: "Choosing" dense subsets of compact sets in plane (or $\mathbb{R}^n$) can be done uniformly as follows: Given $K$ compact, let $\{r_n : n < \omega \}$ list all rational points in plane. For each $n$, let $d_n$ be the distance of $r_n$ from $K$. If $d_n = 0$, let $x_n = r_n$, otherwise choose $x_n \in K$ to be the unique point in $K$ whose distance from $r_n$ is $x_n$ and which makes the smallest angle (in $[0, 2\pi)$) with the horizontal axis. Now check that $\{x_n : n < \omega\}$ is dense in $K$.
A: You don't need the full axiom of choice to show that a countable union of countable sets is countable. Let $\varphi_n : E_n \to \mathbb N$ be a bijection (you can pick such bijections using countable choice, that is, a product of countably many non-empty sets is non-empty, which allows you to pick a bijection for each $n \in \mathbb N$), and let 
$$
\coprod_{n \ge 1} E_n \to \mathbb N^2, \qquad (n, x) \mapsto (n, \varphi_n(x)).
$$
(the symbol $\coprod$ denotes disjoint union). This is a bijection, and you can easily compute a bijection of $\mathbb N^2$ to $\mathbb N$. The inclusions $E_n \to \bigcup_{n \ge 1} E$ can be put together to give a map $\coprod_{n \ge 1} E_n \to E$ which is surjective, showing that $|E| \le \left| \coprod_{n \ge 1} E_n \right| = |\mathbb N|$. (If you want to define $\le$ using only injections, you can use the Cantor-Bernstein-Schröder Theorem, which does not rely on the axiom of choice.)
Hope that helps,
