Set Closure is Smallest Closed Set/Closure Operator

Theorem
Let $S$ be a set.

Let $\operatorname{cl}: \mathcal P(S) \to \mathcal P(S)$ be a closure operator.

Let $T \subseteq S$.

Then $\operatorname{cl} (T)$ is the smallest closed set (with respect to $\operatorname{cl}$) containing $T$ as a subset.

Proof
By Closure is Closed, $\operatorname{cl} (T)$ is closed.

Suppose that $C$ is closed and $T \subseteq C$.

By the definition of closure operator, $\operatorname{cl}$ is $\subseteq$-increasing, so:
 * $\operatorname{cl} (T) \subseteq \operatorname{cl} (C)$

Since $C$ is closed, $\operatorname{cl}(C) = C$, so:
 * $\operatorname{cl} (T) \subseteq C$

so $\operatorname{cl}(T)$ is the smallest closed set containing $T$ as a subset.