= Solution
The implication <convergence in probability implies convergence in distribution> follows by comparing nearby <events>. Fix a <continuity> point $x$ of the <cumulative distribution function> $F_X$ and let $\varepsilon>0$. If $X\leq x-\varepsilon$ and $|X_n-X|\leq\varepsilon$, then $X_n\leq x$. Conversely, $X_n\leq x$ and $|X_n-X|\leq\varepsilon$ imply $X\leq x+\varepsilon$. Thus
$$
F_X(x-\varepsilon)-\mathbb P(|X_n-X|>\varepsilon)
\leq F_{X_n}(x)
\leq F_X(x+\varepsilon)+\mathbb P(|X_n-X|>\varepsilon).
$$
The error <probability> tends to zero by <convergence in probability>. Taking lower and upper <limits> and then letting $\varepsilon\downarrow0$, <continuity> at $x$ squeezes both bounds to $F_X(x)$. This is exactly <convergence in distribution>.
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