Derived Set in T1 Space is Closed

Theorem
Let $\left({X, \tau}\right)$ be a $T_1$ space.

Let $S \subseteq X$.

Let $S'$ be the derived set of $S$.

Then $S'$ is closed.

Proof
Let $x \in S''$.

Let $U$ be an open neighborhood of $x$.

Then by the definition of derived set, $U$ contains an element $y$ of $S'$ such that $x \ne y$.

Then $U \setminus \left\{{x}\right\}$ is an open neighborhood of $y$ by the definition of a $T_1$ space.

The definition of derived set is applied once more to see that $U \setminus \left\{{x}\right\}$ contains an element $z$ of $S$.

Thus $z \in U \cap S$ and $z \ne x$.

As this holds for all open neighborhoods $U$ of $x$, it follows that $x \in S'$.

This holds for all $x \in S''$.

It follows by definition of subset that $S'' \subseteq S'$.

Thus $S'$ is closed.