We denote $\Bbb P_n$ and $\Bbb P$ the measures associated with $X_n$ and $X$ respectively.
Assume that $E[f(X_n)]\to E[f(X)]$ for all $f$ uniformly continuous and bounded. Fix $F$ a closed set and let $O_n:=\{x\in S,d(x,F)<n^{-1}\}$. Then the map $f_n\colon x\mapsto \frac{d(x,O_n^c)}{d(x,O_n^c)+d(x,F)}$ is uniformly continuous and bounded.
Claim: $\limsup_{n\to +\infty}\Bbb P_n(F)\leqslant \Bbb P(F)$.
Indeed, $f_n(x)=1-\frac{d(x,F)}{d(x,O_n^c)+d(x,F)}$ is monotone, bounded by $1$ and converges pointwise to the characteristic function of $F$. We have for each $n$ and $N$,
$$\Bbb P_n(F)\leqslant \int f_N(x)d\Bbb P_n,$$
so for all $N$,
$$\limsup_{n\to +\infty}\Bbb P_n(F)\leqslant \int f_N(x)dP,$$
and we conclude by monotone convergence.
Now, fix $f$ a continuous function such that $0\leqslant f\leqslant 1$. Let $F_{n,j}:=\{x\in S,f(x)\geqslant \frac jn\}$.
\begin{align*}
\int_S fd\Bbb P_N-\int_S fd\Bbb P &\leqslant \sum_{k=0}^n\frac kn\left(\Bbb P_N\left(\frac kn\leqslant
f(x)<\frac{k+1}n\right)-\Bbb P\left(\frac kn\leqslant f(x)<
\frac{k+1}n\right)\right)+\frac 1n\\\
&=\sum_{j=0}^n\frac jn\Bbb P_N(F_{n,j})-\sum_{j=1}^{n+1}\frac{j-1}n\Bbb P_N(F_{n,j})
-\sum_{j=0}^n\frac jn\Bbb P(F_{n,j})\\&+\sum_{j=1}^{n-1}\frac{j-1}n\Bbb P(F_{n,j})+\frac 1n\\
&=\frac 1n\sum_{j=1}^n\left(\Bbb P_N(F_{n,j})-\Bbb P(F_{n,j})\right)+\frac 1n.
\end{align*}
Taking $\limsup_{N\to +\infty}$ and doing the same for $1-f$ instead of $f$, we get the wanted result.
It's a part of portmanteau theorem.
A good reference for questions about weak convergence is Billingsley's book Convergence or probability measures.
Best Answer
Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6).
Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem:
Let $\mu_n$, $1\leq n \leq \infty$ be probability measures with characteristic functions $\phi_n$.
(i) If $\mu_n$ converges weakly to $\mu_\infty$, then $\phi_n(t)$ converges pointwise to $\phi_\infty(t)$.
(ii) If $\phi_n(t)$ converges pointwise to a limit $\phi(t)$ that is continuous at $0$, then the associated sequence of distributions $\mu_n$ is tight and converges weakly to a measure $\mu$ with characteristic function $\phi$.
Statement (i) follows from the Portmanteau Theorem characterization of weakly convergence that uses bounded and continuous functions, by noting that $\mathrm e^{itx}$ is bounded and continuous in $x$.
For (ii), Durrett first proves that sequence of distributions $\mu_n$ is tight (not easy, uses the continuity of $\phi$ at $0$; I'll try to update some details later), which in turn implies the existence of a weakly convergent subsequence. The distribution, call it $\mu$, this subsequence converges weakly to, must have characteristic function $\phi$. Moreover, every subsequence has a further subsequence that converges weakly to $\mu$. Using again the Portmanteau Theorem characterization mentioned above (and a general topological fact: if every subsequence has a further subsequence that converges to some point, then the whole sequence converges to that point) one can show that the whole sequence of distributions $\mu_n$ converges weakly to $\mu$.
Edit: The tightness follows from the observation that: $$ \mu_n\left(\{x\;|\;|x|>2u^{-1} \}\right) \leq u^{-1}\int_{-u}^u (1- \phi_n(t))dt,$$ using Fubini's theorem to re-write the integral as: $$ u^{-1}\int_{-u}^u (1- \phi_n(t))dt = \int_{-\infty}^\infty \left(1- \frac{\sin(ux)}{ux}\right)\mu_n(dx),$$ and the fact that $|\sin x|\leq |x|$ for all $x$.
Edit2: For relationship of tightness and weak convergence topology see also Prokhorov's Theorem and its corollaries.