Proof of Equivalent Characterisation of Complete Metric Space without using Compactifications

Question

While studying about Completely Metrizable spaces, I came across this theorem -

For a metric space $X$, the following are equivalent

1. $X$ is completely metrizable
2. $X$ is a $G_\delta$ in its completion $\hat X$
3. $X$ is a $G_\delta$ in every metric embedding
4. $X$ is a $G_\delta$ in $\beta X$
5. $X$ is a $G_\delta$ whenever densely embedded in a Tychonoff space

Now, I do not know about Compactifications. So, I was wondering whether I could just prove $1-3$ and $5$ were equivalent, without going through $4$.

As the book provides proofs for $1\!\!\implies\!\!2\!\!\implies\!\!3\!\!\implies\!\!1$, and $5\!\!\implies\!\!3$, all I need is a proof for one of $1,2,3$ implying $5$. However, I wasn't able to prove this myself, nor could I find it online. So, can we find such a proof?

Answer 1

Suppose that $X$ is a completely metrizable dense subset of a Tikhonov space $Y$. Let $d$ be a compatible complete metric on $X$, and let $M$ denote the complete metric space $\langle Y,d\rangle$. Then the identity map is a homeomorphism of $X$ onto the complete metric space $M$, and in this answer I showed that the existence of such a homeomorphism implies that $X$ is a $G_\delta$ in $Y$. (The ‘Kuratowski’s result’ mentioned in that answer is proved in this answer.)