Show that $cl(A) = int(A^c)^c$ 
Show that $\operatorname{cl}(A) = \operatorname{int}(A^{\complement})^{\complement}$

I already proved one part:
$$\operatorname{int}(A^{\complement}) \subseteq A^{\complement} \implies \operatorname{int} (A^{\complement})^{\complement} \subseteq A \implies \operatorname{int}(A^{\complement})^{\complement} \subseteq \operatorname{cl}(A)$$
I'm having troubles with the other side. We want to show that $\operatorname{cl}(A)^{\complement} \subseteq \operatorname{int}(A^{\complement})$. I've tried:
$$A\subseteq \operatorname{cl}(A) \implies \operatorname{cl}(A)^{\complement} \subseteq A^{\complement}$$
But I can't conclude that $\operatorname{cl}(A)^{\complement}\subseteq \operatorname{int}(A^{\complement})$ since it doesn't have to be true. 
 A: $int(A^c)$ is an open subset contained in $A^c$, thus $(int(A^c))^c$ is a closed subset which contains $A$ thus contains $c(l(A))$.
$cl(A)$ is a closed subset which contains $A$, thus $cl(A)^c$ is an open subset contained in $A^c$ thus $cl(A)^c\subset int(A^c)$ this implies that $int(A^c)^c\subset cl(A)$.
A: $"\subseteq"$: Suppose $p\notin int(A^c)^c, \text{then } p\in int(A^c)$, then there exists a neighborhood $N$ of $p$ such that $p\in N \subseteq A^c$, which implies $p\notin A$ AND $p$ is not a limit point of $A$, so $p\notin \text{cl}(A)$, hence $cl(A)\subseteq int(A^c)^c$
$"\supseteq"$: Suppose $p\notin cl(A)$, then $p\notin A$ AND $p$ is not a limit point of $A$. Since $p$ is not a limit point, there exists a neighborhood $N$ of $p$ such that $N\setminus\{p\} \subseteq A^c$, and since $p\in A^c$, $N\subseteq A^c$, so $p$ is an interior point of $A^c$, so $p\in int(A^c)$. Hence $cl(A) \supseteq int(A^c)^c$. 
A: We will show the desired inclusion by supposing there exists a point in $\operatorname{cl}(A)$ which is not in $\operatorname{int}(A^c)^c$ and reaching a contradiction.
Let $x \in \operatorname{cl}(A)$. Suppose for a contradiction that $x \in \operatorname{int}(A^c)$. Since $x$ is an interior point of $A^c$ there exists a neighborhood $N$ of $x$ such that $N$ is contained in $A^c$. Hence this neighborhood is disjoint from $A$. But if $x$ is in the closure of $A$ then every neighborhood must meet $A$, a contradiction. Therefore $x \not\in \operatorname{int}(A^c)$, i.e. $x \in \operatorname{int}(A^c)^c$. This completes the proof.
