My lecture notes prove the following statement
A complex valued sequence $(z_n)_{n\in\mathbb{N}} = (a_n+ib_n)_{n\in\mathbb{N}}$ is convergent to $L = A+iB \in\mathbb{C}$ if and only if $(a_n)$ and $(b_n)$ are convergent.
The lecturer's proof involves using the Sandwich (Squeeze) Theorem, however I wondered whether there was any particular issue with using a more direct approach, namely:
Proof: Let $(a_n)\to A\in\mathbb{R}$ and $(b_n)\to B\in\mathbb{R}$. Then for all $\epsilon>0$, there exists $N$ such that for all $n>N$, $|a_n - A| < \epsilon/\sqrt{2}$ and $|b_n - B| < \epsilon/\sqrt{2}$. Hence $|z_n - L| = \sqrt{|a_n - A|^2 + |b_n - B|^2} < \sqrt{\epsilon^2/2 + \epsilon^2/2} = \epsilon$. Thus $(z_n)\to L$.
Now suppose $(z_n)$ converges to $L$. Then for any $\epsilon>0$, there exists $N$ such that for all $n>N$,$|z_n -L| < \epsilon$. This implies $$|(a_n - A)+i(b_n-B)|=\sqrt{|a_n - A|^2 + |b_n - B|^2}<\epsilon.$$
Hence $|a_n - A|^2 + |b_n - B|^2<\epsilon^2$. And since $|a_n-A|^2\leq|a_n - A|^2 + |b_n - B|^2<\epsilon^2$, we have that $(a_n)\to A$ and repeating this argument we have, similarly, $(b_n)\to B$. $$\tag*{$\blacksquare$}$$
I wondered whether there was any particular drawback of my proof and whether or not it is even correct. Is there any real reason for using the Sandwich Theorem, or is it simply personal preference? (Since the length of the proofs was not much different).
Your proof is perfect and well explained.
I would say it is a matter of personal preference to choose one or the other proof.