Prove that $X_n\cdot Y_n \to a\cdot X$ in distribution Problem:

Let $X, X_n, Y_n\, (n\in \Bbb{N})$ be random variables such that $X_n \to X$ in distribution and $Y_n \to a$ in probability for some $a\in \Bbb{R}$. Then $$X_n\cdot Y_n \to a\cdot X \text{ in distribution}.$$

My attempt:
If we could show that $Z_n=(X_n, Y_n)\to (X,a)$ in distribution, then we could apply then continuous mapping theorem to $f(x,y) = x\cdot y$ and conclude that $X_n\cdot Y_n = f(X_n, Y_n) \to f(X,a) = X\cdot a$ in distribution. I haven't been able to prove it.
Is there a nice way of seeing this? Is there a better way to go about the proof?
 A: One possible way to check that $Z_n \to Z$ in distribution is to verify (see Portemanteau theorem)
$$ \limsup_n\mathbb{P} [Z_n \in F]  \leq \mathbb{P}[Z \in F] $$
Since $X_n \to X $ in distribution 
$$ \limsup_n \mathbb{P} [\vert X_n\vert \geq A] \leq \mathbb{P}[ \vert X\vert \geq A]$$
Fix $\epsilon>0$ Choose $A$ large, so that $\mathbb{P}[\vert X\vert \geq A]\leq \epsilon $
Now consider $N_0$ so that $$n \geq N_0 \Rightarrow \mathbb{P} [\vert Y_n - a\vert \geq \gamma] \leq \epsilon$$ (Here we are using that $Y_n \to a$ in probability)
$$n \geq N_0 \Rightarrow \mathbb{P} [\vert X_n \vert \geq A] \leq \epsilon$$ 
 Then
\begin{align*}
\mathbb{P} [X_n Y_n \in F] &\leq \mathbb{P} [\vert X_n \vert \geq A] + \mathbb{P} [X_n Y_n \in F,\vert X_n \vert\leq A]\\
&  \leq \mathbb{P} [\vert X_n \vert \geq A] + \mathbb{P} [X_n a \in F^{A\gamma} ,\vert X_n \vert\leq A ] + \mathbb{P} [\vert Y_n  - a\vert \geq \epsilon]\\
&  \leq \epsilon + \mathbb{P} [X_n a \in F^{A\gamma}]  + \epsilon\\
&  \leq \epsilon + \mathbb{P} [X a \in \overline{F^{A\gamma}}] + \epsilon+ \epsilon
\end{align*}
Now for small enought $\gamma$ one has that $\mathbb{P} [X a \in \overline{F^{A\gamma}}] \leq \mathbb{P} [X a \in F] + \epsilon$
choose such $\gamma$
putting the pieces together  one finds that $\forall \, \epsilon$ $\exists N_0$ such that for $n> N_0$:
$$
\mathbb{P} [X_n Y_n \in F] \leq  \mathbb{P} [X a \in F] + 4\epsilon$$
that is 
$$ \limsup_n\mathbb{P} [X_n Y_n \in F]  \leq \mathbb{P}[X a \in F] + 4 \epsilon $$
since $\epsilon>0$ is arbitrary
$$ \limsup_n\mathbb{P} [X_n Y_n \in F]  \leq \mathbb{P}[X a \in F] $$
and therefore $ X_n Y_n \to X a$ in distribution
remark: the set $ F^{A\gamma}:= \{z : d(z,F) < A\gamma \}$ 
Hope this helps
