$\newcommand{\R}{\mathbb{R}}
\newcommand{\ep}{\varepsilon}
\newcommand{\p}{\partial}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}$
This is to try to provide a simplification and detalization of the answer by Mateusz Kwaśnicki.

Suppose that the distribution of $X$ is absolutely continuous (with respect to the Lebesgue measure) and $X_n\to X$ in distribution. We are going to show that then $X_n\to X$ in in convex distance, that is,
$$\sup_K|\mu_n(K)-\mu(K)|\overset{\text{(?)}}\to0$$
(as $n\to\infty$), where $\mu_n$ and $\mu$ are the distributions of $X_n$ and $X$, respectively, and $\sup_K$ is taken over all measurable convex sets in $\R^d$.

Take any real $\ep>0$. Then there is some real $R>0$ such that $P(X\notin Q_R)\le\ep$, where $Q_R:=(-R/2,R/2]^d$, a left-open $d$-cube. Since $X_n\to X$ in distribution and $P(X\in\p Q_R)=0$, there is some natural $n_\ep$ such that for all natural $n\ge n_\ep$ we have
$$P(X_n\notin Q_R)\le2\ep.$$
Take a natural $N$ and partition the left-open $d$-cube $Q_R$ naturally into $N^d$ left-open $d$-cubes $q_j$ each with edge length $\de:=R/N$, where $j\in J:=[N^d]:=\{1,\dots,N^d\}$.

Using again the conditions that $X_n\to X$ in distribution and $\mu$ is absolutely continuous (so that $\mu(\p q_j)=0$ for all $j\in J$), and increasing $n_\ep$ is needed, we may assume that for all natural $n\ge n_\ep$
$$\De:=\sum_{j\in J}|\mu_n(q_j)-\mu(q_j)|\le\ep.$$

Take now any measurable convex set $K$ in $\R^d$. Then
$$|\mu_n(K)-\mu(K)|\le|\mu_n(K\cap Q_R)-\mu(K\cap Q_R)| \\
+|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|
$$
and
$$|\mu_n(K\setminus Q_R)-\mu(K\setminus Q_R)|\le
P(X_n\notin Q_R)+P(X\notin Q_R)\le3\ep.
$$

So, without loss of generality (wlog) $K\subseteq Q_R$. Let
$$J_<:=J_{<,K}:=\{j\in J\colon q_j\subseteq K^\circ\},$$
$$J_\le:=J_{\le,K}:=\{j\in J\colon q_j\cap \bar K\ne\emptyset\},$$
$$J_=:=J_{=,K}:=\{j\in J\colon q_j\cap\p K\ne\emptyset\},$$
where $K^\circ$ is the interior of $K$ and $\bar K$ is the closure of $K$.

The key to the whole thing is

**Lemma.** $|\bigcup_{j\in J_=}q_j|\le2d(d+2)R^{d-1}\de$, where $|\cdot|$ is the Lebesgue measure.

This lemma will be proved at the end of this answer. Using the absolute continuity of the distribution of $X$, we can take $N$ so large that for any Borel subset $B$ of $\R^d$ we have the implication
$$|B|\le2d(d+2)R^{d-1}\de\implies \mu(B)\le\ep.$$

Using now the lemma, for $n\ge n_\ep$ we have
$$\mu_n(K)-\mu(K)
\le\sum_{j\in J_\le}\mu_n(q_j)-\sum_{j\in J_<}\mu(q_j) \\
\le\sum_{j\in J_\le}|\mu_n(q_j)-\mu(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep.
$$
Similarly,
$$\mu(K)-\mu_n(K)
\le\sum_{j\in J_\le}\mu(q_j)-\sum_{j\in J_<}\mu_n(q_j) \\
\le\sum_{j\in J<}|\mu(q_j)-\mu_n(q_j)|
+\mu
\Big(\bigcup_{j\in J_=}q_j\Big)\le\De+\ep\le2\ep.
$$
So, $|\mu_n(K)-\mu(K)|\le2\ep$. That is, the desired result is proved modulo the lemma.

*Proof of the lemma.* Since $K$ is convex, for any $x\in\p K$ there is some unit vector $\nu(x)$ such that $\nu(x)\cdot(y-x)\le0$ for all $y\in K$ (the support half-space thing), where $\cdot$ denotes the dot product. For each $j\in[d]$, let
$$S_j^+:=\{x\in\p K\colon\nu(x)_j\ge1/\sqrt d\},\quad
S_j^-:=\{x\in\p K\colon\nu(x)_j\le-1/\sqrt d\},$$
$$J_{=,j}^+:=\{j\in J\colon q_j\cap S_j^+\ne\emptyset\},\quad
J_{=,j}^-:=\{j\in J\colon q_j\cap S_j^-\ne\emptyset\},$$
where $v_j$ is the $j$th coordinate of a vector $v\in\R^d$.
Note that
$\bigcup_{j\in[d]}(S_j^+\cup S_j^-)=\p K$ and hence
$\bigcup_{j\in[d]}(J_{=,j}^+\cup J_{=,j}^-)=J_=$, so that
$$\Big|\bigcup_{j\in J_=}q_j\Big|
\le
\sum_{j\in[d]}\Big(\Big|\bigcup_{j\in J_{=,j}^+}q_j\Big|+\Big|\bigcup_{j\in J_{=,j}^-}q_j\Big|\Big)
\le\de^d
\sum_{j\in[d]}(|J_{=,j}^+|+|J_{=,j}^-|),\tag{*}
$$
where now $|J_{=,j}^\pm|$ denotes the cardinality of $J_{=,j}^\pm$.

Now comes the key step in the proof of the lemma:
Take any $x$ and $y$ in $S_d^+$ such that $x_d\le y_d$. We have the support "inequality" $\nu(x)\cdot(y-x)\le0$, which implies
$$\frac{y_d-x_d}{\sqrt d}\le\nu(x)_d(y_d-x_d)\le\sum_{j=1}^{d-1}\nu(x)_j(x_j-y_j)
\le|P_{d-1}x-P_{d-1}y|,
$$
where $P_{d-1}x:=(x_1,\dots,x_{d-1})$. So, we get the crucial Lipschitz condition
$$|y_d-x_d|\le\sqrt d\,|P_{d-1}x-P_{d-1}y| \tag{**}$$
for all $x$ and $y$ in $S_d^+$.

Partition the left-open $(d-1)$-cube $P_{d-1}Q_R$ naturally into $N^{d-1}$ left-open $(d-1)$-cubes $c_i$ each with edge length $\de=R/N$, where $i\in I:=[N^{d-1}]$.
For each $i\in I$, let
$$ J_{=,d,i}^+:=\{j\in J_{=,d}^+\colon P_{d-1}q_j=c_i\},\quad
s_i:=\bigcup_{j\in J_{=,d,i}}q_j,
$$
so that $s_i$ is the "stack" of all the $d$-cubes $q_j$ with $j\in J_{=,d}^+$ that $P_{d-1}$ projects onto the same $(d-1)$-cube $c_i$. Let $r_i$ be the cardinality of the set $J_{=,d,i}$, that is,
the number of the $d$-cubes $q_j$ in the stack $s_i$. Then for some two points $x$ and $y$ in $s_i\cap S_d^+$ we have $|y_d-x_d|\ge(r_i-2)\de$, whence, in view of (**),
$$\sqrt d\,\sqrt{d-1}\,\de\ge\sqrt d\,|P_{d-1}x-P_{d-1}y|\ge|y_d-x_d|\ge(r_i-2)\de,$$
so that $r_i\le d+2$. So,
$$|J_{=,d}^+|=\sum_{i\in I}r_i\le\sum_{i\in I}(d+2)=(d+2)N^{d-1}=(d+2)(R/\de)^{d-1}.$$
Similarly, $|J_{=,j}^\pm|\le(d+2)(R/\de)^{d-1}$ for all $j\in[d]$. Now the lemma follows from (*).