Does proving that closed subset of Polish space is Polish require axiom of countable choice?

Question

Let $C$ be a closed subset of polish space $P$. It is trivial that $C$ is also completely metrizable, but how do we prove that $C$ is separable? I came up with this method: We can prove that separable metrizable space is second countable without $\mathsf{AC}_\omega$. And then we can prove that $C$ is also second countable, hence it is separable (!). But I've heard that second countable spaces are separable is equivalent to $\mathsf{AC}_\omega$: Second-countable implies separable/Axiom countable choice. Maybe there is a way to avoid from using $\mathsf{AC}_\omega$?

Since $P$ is Polish space we can construct a choice function for the entire family of nonempty closed sets without $\mathsf{AC}_\omega$: Constructing a choice function in a complete & separable metric space. Let $D$ be a countable dense subset of $P$. Then $\{B(x, 1/n) \mid x \in D, n \in \mathbb{N}_{>0}\}$ is a countable basis for $P$, so $\{B(x, 1/n) \cap C \mid x \in D, n \in \mathbb{N}_{>0}\}$ is a countable basis for $C$. Let $f$ be the choice function for the entire family of nonempty closed sets of $P$ by above question. If $B(x, 1/n) \cap C \neq \varnothing$, Let $m = \min\{m \in \mathbb{N}_{>0} \mid \overline{B}(x, 1/n-1/m) \cap C \neq \varnothing\}$. Then we can pick $x_0 = f(\overline{B}(x, 1/n-1/m) \cap C)$ from each elements of countable basis, so we can obtain a countable dense subset of $D$. Is this correct?

Answer 1

Yes. The argument is fine. Since $C$ is closed, simply consider $C_{n,m}=C\cap \overline{B}(x_n,1/m)\mid n<m\in\Bbb N\}$ and choose from these sets.

Of course we rely on the fact that $C$ is closed to prove that the intersection is closed, and if $C$ is an arbitrary set, this is no longer the case. And indeed, it is possible that a subspace of $\Bbb R$ is not separable, even though it is completely metrizable.