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

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

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

From Set is Subset of its Topological Closure:
 * $H \subseteq \operatorname{cl} \left({H}\right)$

We have that Limit Point of Subset is Limit Point of Set.

But $x$ is not a limit point of $\operatorname{cl} \left({H}\right)$.

So by the Rule of Transposition, $x$ cannot be a limit point of $H$.

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