First notice that $Y_n$ is a non-negative number, so $\operatorname{E}[Y_n]=Y_n$, then we have that $Y_n\to 0$. Now let $P_n$ the probability measure induced by $X_n$, and $P_X$ the probability measure induced by $X$ and set $Q:=\frac1{2}P_X+\sum_{n\geqslant 1}\frac1{2^{n+1}}P_n$, then its easy to check that $Q$ is a probability measure and that $P_n\ll Q$ and $P_X\ll Q$, therefore there are Radon-Nikodym derivatives $f_n$ and $f_X$ such that $P_n=f_n\cdot Q$ and $P_X=f_X\cdot Q$.
Let $\mathcal{B}(\mathbb{R})$ the Borel $\sigma $-algebra of $\mathbb{R}$ and note that
$$
\begin{align*}
d_{TV}(X_n,X)&=\sup_{A\in \mathcal{B}(\mathbb{R})}\left|\int_{A}(f_n-f_X)dQ\right|\\
&=\int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ\\
&=\frac1{2}\int_{\mathbb{R}}|f_n-f_X|dQ
\end{align*}
$$
where the last equality follows from the fact that
$$
\int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ=\int_{\{f_n-f_X< 0\}}(f_X-f_n)dQ
$$
as $\int_{\mathbb{R}}(f_n-f_X)dQ=0$. Therefore $f_n \xrightarrow{L_1}f_X$, what implies that $X_n\xrightarrow{\text{dist.}}X$ (as $L_1$ convergence implies weak convergence of measures).∎
An easier way to state the same is that for all $c\in \mathbb{R}$
$$
|P_n((-\infty ,c])-P_X((-\infty ,c])|\leqslant d_{TV}(X_n,X)\to 0
$$
so $X_n\xrightarrow{\text{dist.}}X$.
Separable (but not complete) metric space with a measure that is not tight.
Perhaps Murthi is right, that there is no simple example. So I will provide a not-so-simple example.
Let $B \subseteq [0,1]$ be a Bernstein set. This means:
$$
\text{for every uncountable closed set $F \subseteq [0,1]$, we have both $F\cap B \ne \varnothing$ and $F\setminus B \ne \varnothing$.}
\tag1$$
(A Bernstein set $B$ exists according to ZFC.) Our separable metric space is $B$ with the restriction of the usual metric $|x-y|$ of $[0,1]$.
Let $\lambda^*$ Lebesgue outer measure on $[0,1]$. The Bernstein set $B$ has $\lambda^*(B) = 1$, so defining $\mu$ by
$$
\mu(E) = \lambda^*(E)\quad\text{for all } E \in \operatorname{Borel}(B)
\tag2$$
gives us a probability measure $\mu$ on $B$. But $\mu$ is not tight ... Indeed, let $K \subseteq B$ be compact. Then by $(1)$, $K$ is countable, so by $(2)$, $\mu(K) = 0$.
Best Answer
Your question is related to the concepot of a universally measurable metric space. All definitions and Theorems below are taken from the book "Real Analysis and Probability" by R.M. Dudley.
A separable metric space $(S,d)$ is universally measurable if for every (Borel) probability measure $P$ on the completion $T$ of $S$, there are $A \subset S \subset B$ with $A,B \subset T$ Borel measurable and with $P(A) = P(B)$, see the Definition in $11.5.
One can then show (cf. Theorem 11.5.1) that a separable metric space $(S,d)$ is universally measurable iff every (Borel) probability measure on $S$ is tight.
Finally (cf. Theorem 11.5.3) a result of Le Cam states that if $(S,d)$ is a metric space and if $P_n$ is tight for every $n$ and $P_n \to P_0$, then $(P_n)_n$ is uniformly tight.
In particular, if $(S,d)$ is separable and universally measurable, then every converging sequence of (Borel) probability measures is uniformly tight.
Hence, your result holds if your space under consideration is universally measurable. One condition when this holds is if $X$ is a Borel subset of its completion.
But if $X$ is not universally measurable, then (by Theorem 11.5.1 mentioned above), there is a law $P$ on $X$ which is not tight. Now, if you set $P_n =P$ for all $n$, then $P_n \to P$, but $(P_n)_n$ is not (uniformly) tight.
Thus, your desired result holds iff $X$ is universally measurable.