$\textbf{Hint:}$ Use $||a|-|b|| \leq |a-b|$ in the $\epsilon-\delta$ analysis.
It's been awhile since I've studied any kind of analysis, I'm not sure if I'm on the right track. I want to show what I've done and see if it makes sense so far.
I need to show that for any $\epsilon$, I can find a $\delta$ such that $|a-b| < \delta \implies ||a|-|b|| < \epsilon$.
Using the hint, I break the problem into three cases:
Case (i) when $||a|-|b|| = |a-b|$, case (ii) when $||a|-|b|| < |a-b| < \epsilon$, case (iii) when $||a|-|b|| < \epsilon < |a-b|$.
For case (i), I let $\delta = \epsilon$. Then $|a-b| < \delta \implies ||a|-|b|| < \epsilon$.
For case (ii), I let $\delta = \frac{|a-b|+\epsilon}{2}$. Then $|a-b| < \delta \implies ||a|-|b|| < \epsilon$ (since $||a|-|b|| < |a-b| < \delta < \epsilon$) in this case.
I am not sure about the rest. Any hints appreciated.
No need for cases. Let $\epsilon >0$ and $\delta =\epsilon$. Then if $c\in \mathbb{C}$ and $|z-c|<\delta$, by the reverse triangle inequality, $$||z|-|c||\leq |z-c|<\delta=\epsilon,$$ hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.