Prove if $\displaystyle\sum_{n=1}^\infty|a_n|<\infty$, then $\displaystyle\left|\sum_{n=1}^\infty a_n\right|\le \sum_{n=1}^\infty\left|a_n\right|$.
Upon initial observation, it seems like we should utilize the Triangle Inequality.
My initial thoughts:
PROOF: Since $\displaystyle \sum_{n=1}^\infty |a_n|$ is convergent and thus bounded, we then know the sequence of its partial sums is bounded (and that $\displaystyle \sum_{n=1}^\infty a_n$ converges since it converges absolutely, implying that its sequence of partial sums is also convergent and bounded). So, $$\left|\sum_{n = 1}^\infty a_n\right| = |a_1 + a_2 + a_3 + \cdots| \le |a_1| + |a_2| + |a_3| + \cdots = \sum_{n=1}^\infty|a_n|$$
Are we allowed to use the Triangle Inequality because the two series' sequences of partial sums are bounded? Or can we just use the Triangle Inequality straightaway?
No, it is not "allowed" to just use the triangle inequality like that. Indeed, you are using the very fact you want to prove, in your proof... The last inequality you wrote using "$\dots$" is exactly what you are trying to prove! This is called a circular proof and it's not really a proof at all.
Instead you need to be more careful. Let $S_N = \sum_{n=1}^N a_n$ be the sequence of partial sums. Using the actual triangle inequality and induction, you can prove that for all $N$, $$|S_N| \le \sum_{n=1}^N |a_n|.$$
Moreover, it is clear that $\sum_{n=1}^N |a_n| \le \sum_{n=1}^\infty |a_n|$, because the limit of a nondecreasing sequence is bigger than each individual term. Therefore you get that for all $N$, $$|S_N| \le \sum_{n=1}^\infty |a_n|.$$
Now if a sequence $u_n$ converges and for all $n$ you have $|u_n| \le C$, then $\lim_{n \to \infty} |u_n| \le C$. You can apply that to the sequence $\{S_N\}$ and $C = \sum_{n=1}^\infty |a_n|$ to finally get: $$\lim_{N \to \infty} |S_N| = \left| \sum_{n=1}^\infty a_n \right| \le \sum_{n=1}^\infty |a_n|.$$