Polish Spaces and the Hilbert Cube

I take it that you've already normalized $d$ such that $0 \leq d \leq 1$ (otherwise replace $d$ by $\frac{d}{1+d}$).

As you've said, the function $f: x \mapsto f(x) = (d(x,x_{n}))_{n \in \mathbb{N}}$ is continuous and injective. Let $f(y_{m}) \to f(y)$ be a convergent sequence in $f(X)$. We want to show that $y_{m} \to y$.

By definition of the product topology, we have $d(y_{m},x_{n}) \xrightarrow{m \to \infty} d(y,x_{n})$ for all $n$. Let $\varepsilon > 0$ and pick a point $x_{n}$ such that $d(y,x_{n}) < \varepsilon/3$ by density. Since $d(y_{m},x_{n}) \to d(y,x_{n})$, there is $M$ such that $|d(y_{m},x_{n}) - d(y,x_{n})| < \varepsilon /3$ for all $m \geq M$, so $d(y_{m},x_{n}) < 2 \varepsilon /3$. But then $d(y_{m},y) \leq d(y_{m},x_{n}) + d(x_{n},y)< \varepsilon$ and hence $y_{m} \to y$.

Why is the image a $G_{\delta}$-set? This seems to be much more difficult. I don't see any easier way than to essentially re-prove two classical results on metric spaces which are much more interesting, so I prefer to explain this:

Theorem (Kuratowski) Let $A \subset X$ be a subset of a metrizable space and let $g: A \to Y$ be a continuous map to a completely metrizable space $Y$. Then $g$ can be continuously extended to a $G_{\delta}$-set containing $A$.

Fix a bounded and complete metric on $Y$. For the proof we need the notion of oscillation of $g$ at a point $x \in \overline{A}$ (the closure of $A$ in $X$) defined by $$\displaystyle \operatorname{osc}_{g}(x) = \inf\{\operatorname{diam}g(U \cap A)\,:\, x \in U, \;U\; \text{open}\}. $$ The set $B = \{x \in \overline{A}\,:\,\operatorname{osc}_{g}(x) = 0\}$ is a $G_{\delta}$-set. To see this, note that $B_{n} = \{x \in \overline{A} \,:\, \operatorname{osc}_{g}(x) < \frac{1}{n}\}$ is an open subset of the closed set $\overline{A}$ and $B = \bigcap_{n \in \mathbb{N}} B_{n}$. The continuity of $f$ implies that $A \subset B$. Now define $f: B \to Z$ by $f(x) = \lim g(x_{n})$, where $x_{n} \to x$. It is not hard to show that $f$ is well-defined (because $\operatorname{osc}_{g}(x) = 0$ implies that $g(x_{n})$ is a Cauchy-sequence) and clearly $f$ extends $g$ and is continuous.

The second ingredient we need is:

Theorem (Lavrentiev) Let $X$ and $Y$ be completely metrizable spaces and let $g: A \to B$ be a homeomorphism from $A \subset X$ onto $B \subset Y$. Then there exist $G_{\delta}$-sets $G \supset A$ and $H \supset B$ and a homeomorphism $f: G \to H$ extending $g$.

Let $h = g^{-1}$. Choose $G_{\delta}$-sets $G' \supset A$ and $H' \supset B$ and continuous extensions $g': G' \to Y$ and $h': H' \to X$ by Kuratowski's theorem. Let $Z = \operatorname{graph}(g') \cap \widetilde{\operatorname{graph}}(h') \subset X \times Y$ be the intersection of the graphs (the tilde indicates the 'switch' $\widetilde{(y,x)} = (x,y)$ of coordinates) and let $G = \operatorname{pr}_{X} (Z)$ and $H = \operatorname{pr}_{Y}(Z)$. Obviously, $f = g'|_{G}$ is a homeomorphism of $G$ onto $H$. One can check that $H$ (and thus also $G$ by symmetry) is a $G_{\delta}$-set as follows: The graph of $g'$ is closed in $G' \times Y$ and thus it is a $G_{\delta}$-set and $H$ is its preimage under the continuous map $y \mapsto (h'(y),y)$.

Corollary. If $Y$ is a completely metrizable space and $X \subset Y$ a completely metrizable subspace then $X$ is a $G_{\delta}$-set.

By Lavrentiev's theorem, the inclusion $X \subset Y$ extends to a homeomorphism onto its image.

A further corollary of these ideas is that a subset of a Polish space is Polish if and only if it is a $G_{\delta}$.

More detailed information can be found in any decent book on descriptive set theory, for instance Kechris, Classical descriptive set theory, or Srivastava, A course on Borel sets, both appeared in the Springer Graduate Texts in Mathematics series.