= Solution
Assume first that $X_n\xrightarrow dX$. Because all variables take values in the compact interval $[-C,C]$, each monomial $x^k$ agrees there with a bounded continuous function on $\mathbb R$. The <bounded moment criterion for weak convergence> in this case begins with
$$
\mathbb E[X_n^k]\longrightarrow\mathbb E[X^k]
$$
for every natural number $k$.
Conversely, suppose all moments converge. Let $f$ be bounded and continuous. By the <Weierstrass approximation theorem>, for every $\varepsilon>0$ there is a polynomial $P$ with
$$
\sup_{|x|\leq C}|f(x)-P(x)|\leq\varepsilon.
$$
Moment convergence implies $\mathbb E[P(X_n)]\to\mathbb E[P(X)]$, while
$$
|\mathbb E[f(X_n)-P(X_n)]|leq\varepsilon,
\qquad
|\mathbb E[f(X)-P(X)]|\leq\varepsilon.
$$
Taking the limit superior and then $\varepsilon\downarrow0$ proves $\mathbb E[f(X_n)]\to\mathbb E[f(X)]$, which is <convergence in distribution>.
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