What you’ve suggested doesn’t really make much sense, I’m afraid. You’ve no reason to think that any of the terms $x_n$ is $0$, so in forming the product $x_ny_n$ you’re not in general multiplying $y_n$ by $0$. Go back to the definitions.
- What does it mean to say that $\langle x_n:n\in\Bbb N\rangle\to 0$?
- What does it mean to say that $\langle y_n:n\in\Bbb N\rangle$ is bounded?
- And what does it mean to say that $\langle x_ny_n:n\in\Bbb N\rangle\to 0$?
The answers:
- For each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|x_n|<\epsilon$ whenever $n\ge n_\epsilon$.
- There is a real number $M>0$ such that $|y_n|\le M$ for all $n\in\Bbb N$.
- For each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x_ny_n|<\epsilon$ whenever $n\ge m_\epsilon$.
So now we know what we need to do: we need to show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x_ny_n|<\epsilon$ whenever $n\ge m_\epsilon$. Thus, we need to begin by letting $\epsilon>0$; then we’ll try to use what we’ve been given to find the desired $m_\epsilon$.
We do know that there is some $n_\epsilon$ such that $|x_n|<\epsilon$ whenever $n\ge n_0$; that’s a start on making $|x_ny_n|<\epsilon$. Of course it only works if $|y_n|\le 1$, and we don’t know that that’s true. (It may not be.) We do know, however, that $|y_n|\le M$, so we can at least conclude that $|x_ny_n|\le|x_n|M<\epsilon M$ whenever $n\ge n_0$.
Aha! If we had originally arranged to make $|x_n|<\dfrac{\epsilon}M$, we could now conclude that $|x_ny_n|\le|x_n|M<\frac{\epsilon}M\cdot M=\epsilon\;.$ Can we arrange that? Sure: $\dfrac{\epsilon}M>0$, so by hypothesis there is an $n_{\epsilon/M}\in\Bbb N$ such that $|x_n|<\dfrac{\epsilon}M$ whenever $n\ge n_{\epsilon/M}$. Let $m_\epsilon=n_{\epsilon/M}$. Then
$|x_ny_n|\le|x_n|M<\frac{\epsilon}M\cdot M=\epsilon$
whenever $n\ge m_\epsilon$. We can do this for any $\epsilon>0$, so by definition $\langle x_ny_n:n\in\Bbb N\rangle\to 0$.
It all starts with the definitions: they tell you what you know and what you need to prove.