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.
Completeness doesn't really have anything to do with it. The equivalence in your last paragraph is valid in any topological space as long as the support exists. In particular it is valid in any separable metrizable space, such as $(S, \tau_S)$ for any subset $S \subset X$ whatsoever.
Indeed, let $Y$ be any topological space, $\mu$ a Borel probability measure, and assume that the support of $\mu$ exists; call it $E$. Suppose every open neighborhood of $y \in Y$ has positive measure. Then if $F$ is any closed set not containing $y$, we have that $F^c$ is an open neighborhood of $y$, thus $\mu(F^c) > 0$ and so $\mu(F) < 1$. Hence $F \ne E$. We conclude $y \in E$.
Conversely, suppose $y \in E$ and let $U$ be any open neighborhood of $y$. Then $E \setminus U$ is a proper closed subset of $E$; since $E$ is by definition the smallest closed set of measure $1$, we must have $\mu(E \setminus U) < 1$. Therefore $\mu(U) > 0$.
As an aside, when your set $S$ is open, then $(S, \tau_S)$ actually is Polish - indeed, a subset of a Polish space is completely metrizable iff it is $G_\delta$. See Theorem 3.11 of Kechris, Classical Descriptive Set Theory. However, as noted above, we don't need that here.
Best Answer
Here is a solution that yields a non-atomic Borel measure of full support, assuming that $X$ has no isolated points.
Lemma. Any uncountable Polish space has a non-atomic Borel probability measure.
Proof. It is known that any uncountable Polish space is Borel-isomorphic to $[0,1]$, so it suffices to move Lebesgue's measure over by such an isomorphism. QED
Theorem. Any Polish space $X$ without isolated points admits a non-atomic Borel probability measure of full support.
Proof. Let $\{U_n\}_n$ be a countable basis for the topology of $X$. Viewed as a subspace, each $U_n$ is also a Polish space. Observe that $U_n$ cannot be countable or otherwise by Baire it must have isolated points which would also be isolated points of $X$. Therefore $U_n$ is uncountable and hence by the Lemma there is a non-atomic Borel probability measure $\mu _n$ on $U_n$, which we view as a measure on $X$ by setting $\mu _n(X\setminus U_n)=0$.
To conclude it is then enough to take $$ \mu = \sum_{n=1}^\infty 2^{-n}\mu _n. $$