I want to prove that the set $A_{r,s}(z_0)=\{z\in \mathbb{C} \mid r<|z_0-z|< s\}$ is open. This is my attempt.
Let $z \in A_{r,s}(z_0)$. Then $|z-z_0|-r>0$. Let $r'=[|z-z_0|-r]/2$.
Then clearly $D_{r'}(z) \subset A_{r,s}(z_0)$. Thus $A_{r,s}(z_0)$ is open.
I feel like this answer is not quite right or maybe not right at all. Hope someone could validate this or provide a better proof. Thanks
Basically the idea is correct, but there is still something missing. You actually proved that $$D_{r'}(z) \subseteq A_{r,\infty}(z_0) := \{z \in \mathbb{C}; r<|z-z_0|\},$$ i.e. that $A_{r,\infty}(z_0)$ is open. In fact, it might happen that $x \in D_{r'}(z)$ for $r' := (|z-z_0|-r)/2$, but $|x-z_0| \geq s$.
Hint: Use that $$A_{r,s}(z_0) = A_{r,\infty}(z_0) \cap D_s(z_0).$$