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. 

$int(A^c)$ 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 $int(A^c)^c\subset cl(A)$.
