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We want to show that $\{a_nb_n\}$ is Cauchy, i.e. for any $\epsilon >0$ there exists an integer $N$ such that for integers $m,n$, $n\ge m \ge N$ $$\left| \sum_{k=m}^n a_kb_k\right| \le |M|\sum_{k=m}^n|a_k|<\epsilon$$ Where the second inequality came from the triangle inequality and from the fact that for all $|b_n|< M$ since that sequence is bounded. Since $\sum a$ converges we can make $$\sum_{k=p}^q|a_k|<\frac {\epsilon}{|M|}$$ for two integers $q\ge p\ge Q$. Now pick $N>Q$ and it follows that $$\left| \sum_{k=m}^n a_kb_k\right| \le |M|\sum_{k=m}^n|a_k|<|M|\frac {\epsilon}{|M|}=\epsilon$$ I looked at a solution set on the internet and found it so much more complicated than the proof I gave, leading me to wonder if my proof was wrong.
I know I could not have gotten the bounds or the triangle inequality wrong, but maybe I am not justified in my claim for $Q$. Is that what is wrong with my proof?

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    That's not what is meant by "$\{a_nb_n\}$ is Cauchy." What you want to show is that $c_n=\sum_{i=1}^{n} a_ib_i$ is Cauchy. – 2017-01-12
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    If $\sum_na_n$ does not converge absolutely then $\sum_na_nb_n$ may not converge. – 2017-01-12
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    Specifically, you say "Since $\sum a$ converges, ..." then $\sum |a|$ converges, which is not true. – 2017-01-12
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    Indeed, the statement "For all bounded $b$, $\sum a_ib_i$ converges" is equivalent to "absolute convergence" - that is, $\sum|a_i|$ converges. – 2017-01-12
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    Let $a_n=(-1)^n/n$, and $b_n=(-1)^n$. You will find that $\sum ab$ does not converge. – 2017-01-12

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I think what is wrong is that you assumed $a_n$ to be absolute convergent. In the absolute convergent case, the proof is indeed much simpler.

EDIT: In fact if it is not absolute convergent it does not work at all, for example if you take $a_k=\frac{(-1)^k}{k}$ and $b_k=(-1)^k$, as it was promptly pointed by TonyK (I hastily assumed that the solution you found on the internet was correct)

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    And if not, the proof is impossible. – 2017-01-12
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    Then I am having a problem understand what being Cauchy means. A sum being convergent in $\mathbb{R}^k$ is not equivalent to its sequence of partial sums being Cauchy? – 2017-01-12
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    @Guacho: It is equivalent. But you are showing that $\{a_nb_n\}$ is Cauchy, when you should be showing that $\{\sum_{i\le n}a_ib_i\}$ is Cauchy. As Thomas Andrews already told you. – 2017-01-12
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You will have difficulty convincing the reader of the line \begin{equation*} \sum_{k=p}^q|a_k|<\frac{\epsilon}{|M|} \end{equation*} For example, take $\sum_{k=1}^{\infty} \frac{(-1)^k}{k}$.

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As an addendum: if the series were absolutely convergent, then this would be exactly the discrete $1$-$\infty$ Hölder Inequality