Completely metrizable implies $G_\delta$

It is a consequence of Lavrentyev's theorem that a metrizable space is completely metrizable if and only if it is a $G_\delta$ subset in every completely metrizable space containing it.

In my previous question, Theo and I later discussed something related in the comments and he mentioned that this can be extended.

Question: Suppose $X$ is a completely metrizable space, then for every $Y$ which is $\varphi$, we have that $X$ is a $G_\delta$ subset of $Y$.

We know from the aforementioned that $\varphi$ is at least $X\subseteq Y$ and $Y$ completely metrizable. Can this be generalized further, for example $Y$ is Hausdorff and first countable, etc?

Royden’s Proposition 35 in Section 7.9 is true, but his argument is seriously flawed. The heart of the argument is his Proposition 34, which is essentially this:

Proposition: Let $Y$ be a subset of a topological space $X$, and let $f:Y\to M$ be a continuous map into a complete metric space $(M,d)$. Then $f$ may be extended to a continuous function $\overline f:G\to M$, where $G$ is a $G_\delta$-set in $X$ and $Y \subseteq G$.

His proof seems to be seriously incomplete, in that it requires more than just filling in details; I can’t see any way to make it work without the additional assumption that $Y$ is dense in some $G_\delta$-set in $X$. (In fact it’s false as stated: see below.) In that case one can argue as in the proof of Kuratowski’s result: if $Y$ is a dense subset of $A$, a $G_\delta$ in $X$, the set $G = \{x\in A: \operatorname{osc}_f(x)=0\}$ is also a $G_\delta$ in $X$, and $f$ extends to $G$. In particular, if $Y$ is dense in $X$ we may take $A$ to be $X$ itself. (In this version it’s Theorem 4.3.20 in Engelking.) Fortunately, this is enough to give the desired result.

Theorem: Let $Y$ be a dense subset of a Hausdorff space $X$, and let $h:Y\to M$ be a homeomorphism of $Y$ onto a complete metric space $M$. Then $Y$ is a $G_\delta$-set in $X$.

Proof: Since $Y$ is dense in $X$, the corrected version of the proposition ensures that there are a $G_\delta$-set $G\supseteq Y$ and a continuous $\overline h:G\to M$ extending $h$. Let $f = h^{-1}\circ \overline h:G\to Y$, and let $g = \operatorname{id}_G:G\to G$; clearly $f \upharpoonright Y = g\upharpoonright Y = \operatorname{id}_Y$. The range $G$ is Hausdorff, so $f$ and $g$ agree on a closed subset of $G$ and hence on $G \cap \operatorname{cl}Y = G$. But then $f = g$, so $Y = \operatorname{ran}f = \operatorname{ran}g = G$, and $Y$ is therefore a $G_\delta$ in $X$.

Royden correctly requires $X$ to be Hausdorff, but apparently for the wrong reason: judging by his Exercise 8.30, to which he refers at this point, he thinks that he needs the domain of $f$ and $g$ to be Hausdorff to ensure that they’re identical, rather than the range. Here’s the exercise in question:

Let $A \subset B \subset \overline A$ be subsets of a Hausdorff space, and let $f$ and $g$ be two continuous maps of $B$ into a topological space $X$ with $f(u) = g(u)$ for all $u \in A$. Then $f \equiv g$.

Of course this is false, as may be seen by taking $A = \omega$, $B = \omega+1$, $X = \{0,1\}$ with non-empty open sets $\{0\}$ and $X$, $f$ the constant $0$ function on $\omega+1$, and $g$ the characteristic function of $\{\omega\}$ in $\omega+1$.

To see that Royden’s Proposition 34 is false as stated, let $D$ be a set of power $\omega_1$, and let $p$ and $q$ be two points not in $D$. Let $X = D \cup \{p,q\}$, and topologize $X$ as follows: points of $D$ are isolated, and the basic open nbhds of $p$ ($q$, resp.) are the sets of the form $\{p\} \cup (D \setminus C)$ ($\{q\} \cup (D \setminus C)$, resp.), where $C$ is any countable subset of $D$. Let $Y = \{p,q\}$. Let $M = \{0,1\}$ as a subspace of $\mathbb{R}$ with the usual metric, and define $f:Y\to M$ by $f(p) = 0$ and $f(q) = 1$. Then $f$ is a homeomorphism, but $p$ and $q$ don’t have disjoint nbhds in any $G_\delta$ containing both of them, so $f$ has no continuous extension to such a $G_\delta$.