Under the Axiom of Countable Choice, every $G_{\delta }$ subspace of a complete metric space is completely metrizable

Given: The objects, hypotheses, and choice principles stated above.

Let $(X,\mathcal{T})$ be a topological space (def-topological-space) and let $A\subseteq X$. $A$ is a $G_{\delta }$ set of $X$ when there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ of open subsets of $X$ with $A={\bigcap }_{n\in \mathbb{N}}{V}_n$, and an $F_{\sigma }$ set of $X$ when there is a sequence $(F_n{)}_{n\in \mathbb{N}}$ of closed subsets of $X$ with $A={\bigcup }_{n\in \mathbb{N}}{F}_n$. ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion).

If $X$ is completely metrizable and $U\subseteq X$ is open, then $U$ is completely metrizable in its subspace topology. (Every open subspace of a completely metrizable space is completely metrizable).

Assume the Axiom of Countable Choice. If $(X,d)$ is metrizable and $(Y_n{)}_{n\in \mathbb{N}}$ is a sequence of completely metrizable subspaces of $X$, then ${\bigcap }_n{Y}_n$ is completely metrizable. (Under the Axiom of Countable Choice, a countable intersection of completely metrizable subspaces is completely metrizable).