Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 1

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

Suppose that $x$ is not an isolated point of $H$.

Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$.

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

So by Limit Point of Subset is Limit Point of Set it follows that $x$ is a limit point of $\operatorname{cl} \left({H}\right)$.

But then $x$ cannot be an isolated point of $\operatorname{cl} \left({H}\right)$.

This is a contradiction, therefore every isolated point of $\operatorname{cl} \left({H}\right)$ is also an isolated point of $H$.