Isolated Point of Closure of Subset is Isolated Point of Subset

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

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

If $\operatorname{cl}(H)$ has an isolated point, then that point is also an isolated point of $H$.

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

Then by definition:
 * $\exists U \in \tau: U \cap \operatorname{cl}(H) = \left\{{s}\right\}$

From Set is Subset of Closure, we have $H \subseteq \operatorname{cl}(H)$.

So we obtain $U \cap H \subseteq U \cap \operatorname{cl}(H) = \left\{{s}\right\}$.

Then $U \cap H = \varnothing$ or $U \cap H = \left\{{s}\right\}$, depending on: Is $s \in H$?

From Condition for Point being in Closure and the fact that $s \in \operatorname{cl}(H)$, we deduce that $\forall V \in \tau$ such that $s\in V$, $V\cap H \ne \varnothing$.

In particular for $U$, $U \cap H = \left\{{s}\right\}$, then $s \in H$ is an isolated point in $H$.

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

From Point is Isolated iff not a Limit Point, $s$ is not a limit point of $\operatorname{cl}(H)$.

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

So as $s$ is not a limit point of $\operatorname{cl}(H)$ 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 $H$.