A set is closed iff it equals its closure

Statement

Let $X$ be a topological space and let $S \subseteq X$. Then $S$ is closed in $X$ if and only if $\overline{S} = S$.

Remarks

Proof: if $S$ is closed, then $\overline{S} \subseteq S$ by closure is the smallest closed superset, and $S \subseteq \overline{S}$ always holds. If $\overline{S} = S$, then $S$ is closed because the closure is closed. The dual statement is: $S$ is open if and only if $\operatorname{int} S = S$.

Explanations

L1 human

A closed set already contains all the points that you can approach from inside it. So taking the closure adds nothing.