 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 $\epsilon >0$. If $X\le x-\epsilon$ and $|X_n-X|\le \epsilon$, then $X_n\le x$. Conversely, $X_n\le x$ and $|X_n-X|\le \epsilon$ imply $X\le x+\epsilon$. ThusThe error probability tends to zero by convergence in probability. Taking lower and upper limits and then letting $\epsilon \downarrow 0$, continuity at $x$ squeezes both bounds to $F_X(x)$. This is exactly convergence in distribution.