This is pretty straightforward if you use that

$X_n$ tends to $X$ in probability if, and only if, every subsequence of $X_n$ has a sub(sub)sequence that tends to $X$ a.s.

This lemma follows from:

**Fact 1.** If $X_n$ tends to $X$ a.s., then $X_n$ tends to $X$ in probability.

**Fact 2.** If $X_n$ tends to $X$ in probability, it has a subsequence that tends to $X$ a.s.

**Fact 3.** Let $(a_n)$ be a sequence of real numbers. Then $(a_n)$ converges to $a \in \Bbb R$ if, and only if, every subsequence of $(a_n)$ has a sub(sub)sequence that tends to $a$.

## Application

Let $(X_{\phi(n)}Y_{\phi(n)})$ be a subsequence of $(X_nY_n)$. We need to show that it admits a subsequence converging to $XY$ a.s. Since $X_n$ tends to $X$ in probability, there exists $\psi$ such that $X_{\phi(\psi(n))}$ tends to $X$ a.s. Since $Y_n$ tends to $Y$ in probability, there exists $\chi$ such that $Y_{\phi(\psi(\chi(n)))}$ tends to $Y$ a.s. Now, remark that $X_{\phi(\psi(\chi(n)))}Y_{\phi(\psi(\chi(n)))}$ tends to $XY$ a.s.