Proof-verification: $(x_n)$ bounded and $y_n \to 0$ implies $(x_ny_n)\to 0$ Proof-verification: $(x_n)$ bounded and $y_n \to 0$ implies
  $(x_ny_n)\to 0$


My attempted proof:

Let $(x_n)$ be bounded, hence there exists $A\in \mathbb{R}$ such that $|x_n|\leq A$. Furthermore let $y_n$ be a null-sequence; thus for an arbitrary $\varepsilon &gt;0$ there exisits $N\in \mathbb{N}$ such that $|y_n|&lt;\varepsilon$ for all $n\geq N$.

$$|x_n\cdot y_n|\leq |A\cdot y_n|=|A|\cdot |y_n|&lt;|A|\cdot \varepsilon$$ Since $\varepsilon$ does not depend on $|A|$, one can conclude that the claim is indeed true. $\Box$

is that okay?
If we correct the obvious typo then this is fine; exactly this sort of argument is given all the time.

But it's somewhat "informal", in that the conclusion does not look like what's required by the definition. A more formally correct version would read "Since $y_n\to0$, given $\epsilon&gt;0$  there exists $N$ such that $|y_n|&lt;\epsilon/A$ for all $n&gt;N$..."

So whether it's actually "right" depends on the context. In a research paper, or some other context where the point is to convince the  reader it's true, what you wrote is fine - people do write this sort of thing all the time. In a calculus class it's different - there the point is to convince the reader that you know exactly how to prove it, and in that context you're much better off getting a literal $|x_ny_n|\dots&lt;\epsilon$ at the end.

Note: Of course @pre-kidney is also correct in saying that the last sentence is missing the point; that issue doesn't come up if we change it as suggested.
Yes, your proof is okay. Some other users obviously don't share my view. It is my personal opinion they are being too pedantic. I find it amusing that of all answers, this one of mine is so vehemently downvoted and opted for deletion. C'est la vie.