Prove that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ converges absolutely

Here's an alternative:

Suppose $\sum \limits_{n=1}^{+\infty}\left(\left|a_n\right|\right)$ converges.

It follows that $(|a_n|)_{n\in \Bbb N}\longrightarrow 0$, therefore there exists $k>0$ such that $\forall n\in \Bbb N(|a_n|<k)$.

Thus $\forall n\in \Bbb N\left(|a_n|^2<k|a_n|\right)$.

Since $\sum \limits_{n=1}^{+\infty}\left(k|a_n|\right)$ converges and due to all the sequences involved being of non-negative real numbers, it follows that $\sum \limits _{n=1}^{+\infty}\left(|a_n|^2\right)$ converges and the result follows.

Note that $\left(a_n\right)_{n\in \Bbb N}$ can be a complex sequence too.

The problem with your proof as written is that $n$ is arbitrary, so when you choose $N$ you are not choosing a fixed value - you need a different $N$ for each $n$.

You could approach this by noting that there is an $N$ with $|a_m|\lt 1$ for $m\gt N$, and therefore $|a_m|^2\lt |a_m|$, which sets up a comparison using the absolute convergence.