Does convergence in law imply convergence in convex distance?

Question

For two random variables $X$ and $Y$ taking values in $\mathbb{R}^m$, the convex distance $d_c$ is defined as

$$d_c(X,Y) = \sup_{h} \lvert \operatorname{E}(h(X)) - \operatorname{E}(h(Y)) \rvert,$$

where the supremum is taken over all indicator functions of measurable convex subsets of $\mathbb{R}^m$.

For $m=1$, it is easy to see that $d_c$ coincides with the Kolmogorov distance whenever $X$ and $Y$ are continuous, i.e. we have that $d_c(X,Y)= \sup_{x \in \mathbb{R}} \lvert F_X(x) -F_Y(x) \rvert$, where $F_X$ and $F_Y$ denote the cumulative distribution functions of $X$ and $Y$, respectively.

In particular, if $m=1$, we have that if a sequence $(X_n)$ of continuous real-valued random variables converges to another continuous random variable $Y$ in distribution, then $d_c(X_n,Y) \to 0$ as $n \to \infty$. Does this implication continue to hold if $m \geq 2$?

I could neither find a reference, nor a proof of this myself.

EDIT

If continuity of the random variables is defined as having non-atomic distributions, the implication is false (see the counterexample by Iosif Pinelis below).

If continuity is defined as having CDFs which are absolutely continuous with respect to the Lebesgue measure, the question is still open.

Answer 1

The answer is no. For instance, let $Y$ be uniformly distributed on the unit sphere in $\mathbb R^m$ (not the ball, but the sphere) and, for each natural $n$, let $X_n:=(1+1/n)Y$. Then the distribution of each of the random vectors $Y,X_1,X_2,\dots$ is non-atomic and hence, in particular, the values of this distribution on any singleton set are $0$; also, the cumulative distribution function of this distribution is continuous.

Moreover, $X_n\to Y$ in distribution, but $d_c(X_n,Y)=1$ for all $n$.