TITLE: (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
QUESTION [5 upvotes]: Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.
Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.
Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)
Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.
I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)
But i couldn't. Is there any way to show this?
If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.
Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.
Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.
Please help.
If my argument is wrong, please give me a proper proof.

REPLY [2 votes]: Although it’s essentially equivalent to Asaf’s, I prefer the following proof (in ZF) that every separable, completely metrizable space is Baire: I find it easier to work with dense open sets.
Let $\langle X,d\rangle$ be a complete, separable metric space, let $D=\{x_n:n\in\Bbb N\}$ be dense in $X$, let $\{U_n:n\in\Bbb N\}$ be a family of dense open subsets of $X$, and let $V$ be a non-empty open subset of $X$. For $n\in\Bbb N$ and $x\in X$ let $B(x,n)$ be the open $d$-ball of radius $2^{-n}$ centred at $x$, and let $D(x,n)$ be the closed $d$-ball of radius $2^{-n}$ centred at $x$.
Let $$m(0)=\min\left\{n\in\Bbb N:x_n\in G\cap G_0\right\}$$ and $$k(0)=\min\left\{n\in\Bbb N:D\left(x_{m(0)},n\right)\subseteq G\cap G_0\right\}\;.$$
Given $m(n),k(n)\in\Bbb N$, let $$m(n+1)=\min\left\{i\in\Bbb N:x_i\in B\left(x_{m(n)},k(n)\right)\cap G_{n+1}\right\}$$ and $$k(n+1)=\max\left\{2^{n+1},\min\left\{i\in\Bbb N:D\left(x_{m(n+1)},i\right)\subseteq B\left(x_{m(n)},k(n)\right)\cap G_{n+1}\right\}\right\}\;.$$
For $n\in\Bbb N$ let $y_n=x_{m(n)}$. Then
$$B\left(y_n,k(n)\right)\supseteq D\left(y_{n+1},k(n+1)\right)\supseteq B\left(y_{n+1},k(n+1)\right)\supseteq D\left(y_{n+2},k(n+2)\right)\supseteq\ldots$$
for each $n\in\Bbb N$. In particular, $\{y_i:i\ge n\}\subseteq B(y_n,k(n))\subseteq B(y_n,n)$, so $\langle y_n:n\in\Bbb N\rangle$ is a $d$-Cauchy sequence and therefore converges to some $y\in X$. Clearly $y\in D(y_n,k(n))\subseteq G\cap G_n$ for each $n\in\Bbb N$, so $$y\in G\cap\bigcap_{n\in\Bbb N}G_n\ne\varnothing\;.$$ Thus, $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$, and $X$ is Baire.