Show $(A^o)^c=\overline{A^c}$ 
Show $(A^o)^c=\overline{A^c}$. 

($\rightarrow$) $(A^o)^c\subseteq\overline{A^c}$    
I want to show that $(A^o)^c$ is closed and that $A^c\subseteq (A^o)^c$. Then ($\rightarrow$) follows. Since $A^o$ is open $(A^o)^c$ is closed. Since $A^o\subseteq A$, $A^c\subseteq (A^o)^c$. Done. 
($\leftarrow$) $\overline{A^c}\subseteq (A^o)^c$  
This one is trickier. $(A^o)^c$ is closed and since $A^o\subseteq A\implies$ $A^c\subseteq (A^o)^c$. So  $\overline{A^c}\subseteq (A^o)^c$ since $\overline{A^c}$ is the smallest closed set which $A^c$ fits into?
 A: Your second inclusion is fine. $A^o\subseteq A$, so $A^c\subseteq (A^o)^c$, and $A^o$ is open, so $(A^o)^c$ is closed. $\overline{A^o}$ is the intersection of all closed sets containing $A^o$, and $(A^o)^c$ is one of those closed sets, so $\overline{A^c}\subseteq (A^o)^c$.
The argument for the first inclusion, however, is incorrect. The easiest way to show that $(A^o)^c\subseteq\overline{A^c}$ is to chase points: if $x\in(A^o)^c$, then $x\notin A^o$, so every $\epsilon$-ball centred at $x$ intersects $A^c$, and therefore $x\in\overline{A^c}$.
In fact, you really need only that one argument, since each of the implications goes both ways: for any $x$,
$$\begin{align*}x\in(A^o)^c&\iff x\notin A^o\\
&\iff \text{ every }\epsilon\text{-ball about }x\text{ meets }A^c\\
&\iff x\in\overline{A^c}\;,
\end{align*}$$
so of course $(A^o)^c=\overline{A^c}$.
A: In your antepenultimate question you proved that
$$A^{\circ} = A\setminus\partial A.$$
Therefore,
$$\begin{align*}
A^{\circ} &= A\setminus\partial A\\
A^{\circ} &= A\cap (X-\partial A)\\
(A^{\circ})^c &= (A\cap (X-\partial A))^c\\
(A^{\circ})^c &= (X-A)\cup \partial A\\
(A^{\circ})^c &= A^c \cup \partial(A^c) &&\text{(since }\partial A=\partial A^c\text{)}\\
(A^{\circ})^c &= \overline{(A^c)} &&\text{(since for any }B,\ \overline{B}=B\cup\partial B\text{)}
\end{align*}$$
Again the same Moral: Use previous results when possible!
