The assignment
Let $A \subset \mathbb C$. Show that $z \in A^c$ (the closure of $A$), if and only if there exists a sequence $(z_n)$, so that $z_n \in A$ for all $n \geq 1$ and $z_n \to z$.
The problem
My issue with this assignment is that it's so confusingly written that I'm not sure what I'm supposed to prove or allowed to assume going in either direction. For example, going from left to right I know that I can assume that $z \in A^c$, but I can't really prove anything simply based on that, and there is clearly stuff on the other side of the equivalence that I probably need to complete the proof.
So, what am I supposed to prove and what am I allowed to assume?
Here is a rough outline for you to fill the details in:
Let $z \in A^c$ (the smallest closed subset of $\mathbb C$ that contains $A$). Then we have, for any $n \in \mathbb N$, that $B_{1/n}(z) \cap A \neq \emptyset$ (here $B_\epsilon(a)$ is the open ball with centre $a$ and radius $\epsilon$). (Why is this true?)
We may now pick $z_n \in B_{1/n}(z) \cap A$. Then the sequence $(z_n \mid n \in \mathbb N)$ consists of points in $A$ and converges to $z$. (Again, you need to justify this.)
Conversely, suppose $(z_n \mid n \in \mathbb N)$ is a sequence of points in $A$ that converges to $z$. We need to show that $z \in A^c$. Suppose not. Then there is some open set $O \subseteq \mathbb C$ such that $z \in O$ and $O \cap A = \emptyset$. (Why?) But we have that for sufficiently large $n$, $z_n \in O \cap A$ (Why?). Contradiction!