Proof writing: $\sum_{n=1}^{\infty}| a_n|<\infty $ implies $\sum_{n=1}^{\infty} a_n^2<\infty $. 
Let $\sum_{n=1}^{\infty} a_n $ be an absolutely converging series.  By definition, this means $\sum_{n=1}^{\infty} \lvert a_n\rvert $ converges. We want to show that $\sum_{n=1}^{\infty} a^2_n $ converges absolutely as well. 


Here is what I tried:
Let's say $\sum_{n=1}^{\infty} \lvert a_n^2\rvert $ converges. Let $\epsilon > 0$ be given and there exists $N \in \mathbb{N} $ such that 
$\lvert a^2_{m+1}\rvert+ \lvert a^2_{m+2}\rvert + \dots+ \lvert a^2_{n}\rvert< \epsilon$ for all $ n > m \geq N $. By the triangle inequality, 
$$\lvert a^2_{m+1} + a^2_{m+2} + \dots + a^2_{n}\rvert \leq \lvert a^2_{m+1}\rvert+ \lvert a^2_{m+2}\rvert + \dots + \lvert a^2_{n}\rvert $$
so the Cauchy Criterion guarantees that  $\sum_{n=1}^{\infty} a^2_n $ also converges.

Here is my question:  

Is this correct, or do I need to prove anything else?

 A: No, this is not correct: you’ve proved that if $\sum_{n\ge 1}a_n^2$ converges absolutely, then it converges, which is not at all what you’re supposed to be doing. You have assumed precisely the result that you are supposed to be proving, and you’ve ignored the hypothesis. 
You should be assuming that $\sum_{n\ge 1}a_n$ converges absolutely and using that assumption to prove that $\sum_{n\ge 1}a_n^2$ converges absolutely.
HINT: If $\sum_{n\ge 1}a_n$ converges absolutely, then there is an $m\in\Bbb N$ such that $|a_n|<1$ for each $n\ge m$ (why?), and hence $0\le|a_n^2|<|a_n|$ for each $n\ge m$.
A: Hint: Since $(a_n)$ converges absolutely, $\lim_n\mid a_n\mid=0$. This implies there exists $N$ such that $n>N$ implies that $\mid a_n\mid <1$. This implies that for $n>N$, $\mid a^2_n\mid<\mid a_n\mid$.
A: Brian has already pointed out what is wrong with your reasoning. Here is another approach.
Note that if $\sum_{n=1}^\infty|a_n|<\infty$, then 
$$\lim_{n \to \infty} \frac{a^2_n}{|a_n|} = \lim_{n \to \infty} |a_n| = 0.$$
Then take a look at the limit comparison test.

[Might not be useful for now though]
In real analysis,  this is nothing but a simple case of a more general theorem (see for instance Folland's Real Analysis, Chapter 6)
$$
\ell^p(\mathbb{N})\subset\ell^q(\mathbb{N})
$$
for $0<p<q\leq\infty$.
