Convergence of random variables in metric spaces Let $S$ be a metric space equipped with a distance function $d$, and let $X_n,Y_n$ be sequences of random variables having values from $S$. Suppose that $X_n$ converges in distribution to some random variable $X$, and suppose as well that $d(X_n,Y_n)$ convergence to 0 in probability.
Does it follow that $Y_n$ convergence in distribution to $X$?
 A: One can use portmanteau theorem. Fix a closed set $F$. We have to show that $\limsup_{n\to +\infty}\mu(Y_n\in F)\leqslant \mu(X\in F)$. 
To this aim, define $F^\varepsilon:=\{x\in S, d(x,x')\leqslant \varepsilon\mbox{ for some }x'\in F\}$. Then 
$$\mu(Y_n\in F)\leqslant \mu(X_n\in F^\varepsilon,d(X_n,Y_n)\leqslant \varepsilon)+\mu(d(X_n,Y_n)\gt\varepsilon)\leqslant \\
\leqslant\mu(X_n\in F^\varepsilon)+\mu(d(X_n,Y_n)\gt\varepsilon).$$
Since $F^\varepsilon$ is closed, it follows that by the assumption that $d(X_,Y_n)\to 0$ in probability that 
$$\limsup_{n\to +\infty}\mu(Y_n\in F)\leqslant\mu(X\in F^\varepsilon).$$
Since $F^\varepsilon\downarrow F$, the proof is complete.

An alternative way is to show that the convergence $\mathbb E[f(Y_n)]\to \mathbb E[f(X)]$  holds for any $f\colon S\to\mathbb R$ uniformly continuous and bounded.
A: Here is an attempt (let me know if you feel something is wrong or missing):
Fix any continuous and bounded function $f\colon S\to\mathbb{R}$; the goal is to prove that $\mathbb{E} f(Y_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X)$.
From the assumptions, we know that 
$$
\mathbb{E} f(X_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X) \tag{$\dagger$}
$$
and by continuity of $f$ that $f(X_n) - f(Y_n)$ converges to $0$ in probability $(\ddagger)$; writing
$$
\lvert \mathbb{E} f(Y_n) - \mathbb{E} f(X) \rvert \leq \lvert \mathbb{E}[ f(Y_n) - f(X_n)] \rvert + \lvert \mathbb{E} f(X_n) - \mathbb{E} f(X) \rvert
$$
The first term goes to $0$ with $(\ddagger)$, the second by $(\dagger)$; hence $\mathbb{E} f(Y_n)\xrightarrow[n\to\infty]{} \mathbb{E} f(X)$ for any bounded continuous function $f$, i.e. $Y_n\xrightarrow[n\to\infty]{\mathcal{D}} X$.
