Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 2

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

Let $H \subseteq T$ be a subspace of $T$.

Let $\operatorname{cl} \left({H}\right)$ denote the closure of $H$.

Let $x \in \operatorname{cl} \left({H}\right)$ be an isolated point of $\operatorname{cl} \left({H}\right)$.

Then $x$ is also an isolated point of $H$.

Proof
Let $s \in \operatorname{cl} \left({H}\right)$ be isolated in $\operatorname{cl} \left({H}\right)$.

By definition of isolated point, $s$ is not a limit point of $\operatorname{cl}(H)$.

From the definition of set closure, $\operatorname{cl} \left({H}\right)$ is the union of all isolated points and limit points of $H$.

So as $s$ is not a limit point of $\operatorname{cl} \left({H}\right)$ then it can not be a limit point of $H$.

As $s$ is not a limit point of $H$, it follows that $s$ must be an isolated point of $H$.