Question about proof that $A\subset B \implies \overline{A}\subset \overline{B}$ Look at this simple proof that $A\subset B \implies \overline{A}\subset \overline{B}$:
$\overline{B}$ is a set that contains $B$, therefore it contains $A$. Since it's closed and contains $A$, it must contain $\overline{A}$. 
This made sense for me intuitively, but how do I prove the last sentence?

Since $\overline{B}$ it's closed and contains $A$, it must contain $\overline{A}$.


$\bar C$ is the intersection of the set of all closed sets that have $C$ as a subset. Let $A^*$ be the set of all closed sets that have $A$ as a subset. Let $B^*$ be the set of all closed sets that have $B$ as a subset. If $A\subset B$ then $A^*\supset B^*,$ which implies $\bar A=\cap A^*\subset \cap B^*=\bar B.$
