stuck on proof (help with intuition): set $A$ is open $\iff A$ complement is closed

So to prove $"\implies "$ I go by contradiction:

1. Let $A$ be open and $A^c$ (the complement of $A$) also be open.
2. since $A^c$ is open it does not contain its limit points.
3. let $y$ be a limit point of $A^c$, then $y \in (A^c)^c =A$.
4. since $y$ is a limit point of $A$ complement there must always be an element of $A$ complement in any epsilon ball around $y$.
5. But the thing is that A is open: which means any point in a is interior (that is for any point in a, there is a ball around that point that is entirely contained in A)
6. since $y$ is in $A$, this should have been the case for y too.
7. But we see that there is no ball around $y$ which does not contain an element of $A^c$.
8. contradiction: 7) is contradicting 5) & 6) [end of proof $"\implies "$]

*** The other direction though is tricky:

I want to go by contradiction again: I say

1. By contradiction let $A^c$ be closed and $A$ also be closed.
2. Since $A$ is closed it contains all its limit points.
3. ...*** at this point my brain just stops "seeing" what's up [I can probably look this up but I learn better by discussing stuff so any help is appreciated].