 Solution

Every subset is contained in its closure, giving $\overline{A}\subseteq \overline{\overline{A}}$. On the other hand, $\overline{A}$ is already closed by its definition as an intersection of closed sets. It is therefore one of the closed supersets used when taking its own closure, so $\overline{\overline{A}}\subseteq \overline{A}$. HenceThis is the idempotence of the closure operation.