Convergence of $\operatorname{Var}(X_n)$ if $|X_n| \le 1$ I'm having some trouble understanding a certain hypothetical.

If, as mentioned in the title, $|X_n| \le 1$ and $X_n$ converges to $0$ in probability, does the variance of $X_n$ also converge to $0$?


Let $X_n$ be a sequence of random variables converging to $0$ in probability. Now convergence in probability implies convergence in distribution, i.e
$$ \mathbb{E}[g(X_n)]\to \mathbb{E}[g(0))] $$
for any $g$ continuous and bounded. Let $g(x)=x^2 $ for $|x|\leq 1$ and $g(x)=1$ for $|x|>1$. Note that $g$ is continuous and bounded. Furthermore since $X_n\leq 1$ we have $g(X_n)=X_n^2$ (a.s) .
Now for the convergence of the variance :
\begin{align}
\lim_{n\to \infty}Var[X_n]=&\lim_{n\to \infty}E[X_n^2]-E[X_n]^2
\\
=& \lim_{n\to \infty} E[g(X_n)]-E[X_n]^2
\\
=& E[g(0)]-E[0]=0.
\end{align}
