If $b_n$ is a bounded sequence and $\lim a_n = 0$, show that $\lim(a_nb_n) = 0$

You want to show that the absolute value of $b_na_n$ can be made arbitrarily small. But, there exists a positive constant $M$ such that for all $n$, $|b_n|\leq M$, since the sequence $(b_n)$ is bounded. Thus $|b_na_n|\leq M|a_n|$ for all $n$. Since the sequence $(a_n)$ converges to 0, this is all you should need. The argument is basically the Squeeze Theorem from elementary calculus.

Since the $b_n$ are bounded, there is a constant $C$ such that $-C\lt b_n\lt C$ for all $n$. Fix $n$. Suppose first $a_n\gt 0$. Then $-C a_n< a_nb_n < Ca_n$. Similarly, if $a_n\lt 0$ then $Ca_n < a_n b_n < -C a_n$. In short: $|a_n b_n|\lt C|a_n|$.

Now, for any sequence $c_n$, $c_n\to 0$ iff $|c_n|\to 0$. Right?

But $C|a_n|\to 0$ since $a_n\to 0$. Since $|a_nb_n|\le C|a_n|$, we must also have $|a_n b_n|\to 0$.

(This is just a sketch. Let me know if you need me to clarify something.)

You write *... but since $a_n$ is not bounded...*

Presumably you mean that you have not been explicitly told that the sequence $a_n$ is bounded. That is true; you have not.

However, as it happens, the following holds:

Theorem. Let $c_n$ be any sequence of real numbers. If $c_n$ converges, then it is bounded.

Intuitively, why is this true? Well, for large enough $n$, the values of $c_n$ must be close to the limit $L$; so only the "early" values of $c_n$ can get away from the limit, and since there are only finitely many of them, they can only go so far.

So, try to prove this carefully, so you can conclude that $a_n$ is bounded, so that the theorems you want to apply can be applied.