= Solution
Assume <convergence in probability> and fix $\varepsilon>0$. On the event $|Y_n-Y|\leq\varepsilon$, $Y\leq x-\varepsilon$ implies $Y_n\leq x$, while $Y_n\leq x$ implies $Y\leq x+\varepsilon$. Hence
$$
F_Y(x-\varepsilon)-\mathbb P(|Y_n-Y|>\varepsilon)
\leq F_{Y_n}(x)
\leq F_Y(x+\varepsilon)+\mathbb P(|Y_n-Y|>\varepsilon).
$$
Taking lower and upper limits in $n$ removes the error <probabilities>. If $x$ is a <continuity point> of $F_Y$, letting $\varepsilon\downarrow0$ squeezes both bounds to $F_Y(x)$. Thus
$$
\boxed{Y_n\to Y\text{ in probability}\Longrightarrow Y_n\to Y\text{ in distribution}.}
$$
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