Without any additional assumptions (e.g. on the speed of convergence), the assertion is wrong.
(Counter)Example Let $Z \sim N(0,1)$ be a standard Gaussian random variable. If we set $$X_n := \frac{1}{n} Z \qquad Y_n:= \frac{1}{\sqrt{n}} Z$$ then clearly $X_n \to 0$, $Y_n \to 0$ in probability. The density $p$ of $Z$,
$$p(y) = \frac{1}{\sqrt{2\pi}} \exp \left( - \frac{y^2}{2} \right)$$
is continuous and satisfies $p(0)=1/\sqrt{2\pi}$, and therefore it follows that we can choose $r>0$ such that
$$p(y) \geq \frac{3}{4} \frac{1}{\sqrt{2\pi}} \quad \text{for all $|y| \leq r$}.$$
This implies that the transition density $f_n$ of $X_n$ satisfies
$$f_n(y) = \sqrt{n} p \left( \sqrt{n} y \right) \geq \frac{3}{4} \frac{\sqrt{n}}{\sqrt{2\pi}} \quad \text{for all $|y| \leq r/\sqrt{n}$.}$$
Since
$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \geq \int_{|y| \leq r/\sqrt{n}} (f_n(y))-g_n(y)) \, dy$$
and $\|g_n\|_{\infty} \leq n^{1/4}/\sqrt{2\pi}$, we get
$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \geq \left( \frac{3}{4} \frac{\sqrt{n}}{\sqrt{2\pi}} - \frac{n^{1/4}}{\sqrt{2 \pi}} \right) \frac{2r}{\sqrt{n}} \xrightarrow[]{n \to \infty} \frac{3}{2} \frac{r}{\sqrt{2\pi}}>0$$
and so
$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \to 0$$
does not hold true.
Remark For random variables $X$ and $Y$ with density $f$ and $g$, respectively, the total variation distance is defined as
$$d_{\text{TV}}(X,Y) = \frac{1}{2} \int |f-g|.$$
It is possible to show that
$$d_{\text{TV}}(X,Y) = \sup_{A \in \mathcal{B}(\mathbb{R})} |\mathbb{P}(X \in A)-\mathbb{P}(Y \in A)|.$$
The above (counter)example shows that $X_n \to \mu$, $Y_n \to \mu$ in probability does, in general, not imply $d_{\text{TV}}(X_n,Y_n) \to 0$.
Assume that it is true and let it be that $X_n=X$ and $Y_n=Y$ for every $n$ where $X$ and $Y$ are iid and non-degenerate random variables.
Evidently for every pair $i,j$ the rv's $X_i$ and $Y_j$ are iid.
Then $(X_n,Y_n)=(X,Y)$ for all $n$ so of course we have: $$(X_n,Y_n)\stackrel{d}{\to}(X,Y)\tag1$$
But we also have $X_n\stackrel{d}{\to}X$ and $Y_n\stackrel{d}{\to}X$ so under the assumption we arrive at the conclusion that: $$(X_n,Y_n)\stackrel{d}{\to}(X,X)\tag2$$
Combining $(1)$ and $(2)$ we find that $(X,Y)$ and $(X,X)$ have equal distributions which cannot be true.
We conclude that the assumption is false.
Best Answer
Fix $\delta>0$. Choose $K>0$ such that $P(|Y_n|\le K)\ge 1 -\delta$ and $P(|X_n|\le K)\ge 1 -\delta$ for all $n$. (We can do this, because both sequences are tight.) Let $g(x)=(x\wedge K)\vee(-K)$. Then \begin{align} P(|Y_n - X_n| \ge \varepsilon) &\le P(|Y_n - g(Y_n)| \ge \varepsilon/3) + P(|g(Y_n) - g(X_n)| \ge \varepsilon/3) + P(|g(X_n) - X_n| \ge \varepsilon/3)\\ &\le 2\delta + P(|g(Y_n) - g(X_n)| \ge \varepsilon/3). \end{align} But $g$ is nondecreasing, so $g(Y_n)\ge g(X_n)$, and $g$ is bounded and continuous, so we have $$ E|g(Y_n) - g(X_n)| = E[g(Y_n) - g(X_n)] = E[g(Y_n)] - E[g(X_n)] \to 0. $$ Hence, $g(Y_n)-g(X_n)\to 0$ in $L^1$, and therefore also in probability. Letting $n\to\infty$ in the first inequality gives $$ \limsup_{n\to\infty} P(|Y_n - X_n| \ge \varepsilon) \le 2\delta. $$ Letting $\delta\to0$ shows that $|Y_n-X_n|\to0$ in probability.