Help with proving if $s_n$ converges to $0$ and $x_n$ is a bounded sequence, then $\lim(x_ns_n)=0$ 
Prove: If $s_n$ converges to $0$ and $x_n$ is a bounded sequence, then $\underset{n \to \infty}{\lim}(x_ns_n)=0$

I'm have trouble getting started on this proof.  
Since I know $s_n$ converges to $0$ I feel like I should start using the epsilon definition with $x_n$ 
Since $\underset{n \to \infty}{\lim} s_n=0$
So $\left|x_n-s_n\right| \lt \epsilon \implies (=) \quad \left|x_n\right| \lt \epsilon$
But I'm not sure if the aforementioned steps are correct and where to go from here.

Since ${x_n}$ is a bounded sequence, then for every $N_1\in \mathbb N$  there is M >0 $\in \mathbb R$ such that when n> $N_1$,$|x_n|$< M.We also know that since $lim_{n\rightarrow\infty} s_n$ =$0$, for every $\epsilon$ >0 $\in \mathbb R$,there is an $N_2\in \mathbb N$ such that when n> $N_2$, $|s_n|$< $\epsilon$. Let $\epsilon$'= M$\epsilon$ and choose n > max {$N_1$,$N_2$}. By the Cauchy Schwarz inequality:
$|x_n s_n|\leq$ $|x_n||s_n|$ < M$\epsilon$ = $\epsilon$'.
So $lim_{n\rightarrow\infty} (x_n s_n)$ = 0.
That does it.
Addendum: Andre Nicolas correctly pointed out that ($x_n$) need not itself have a limit even though it's bounded. While this is very true, the argument above is completely general and refers to the values of the product of the sequence ranges. I've clarified some steps above to show this. 
