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

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 $\operatorname{cl} \left({H}\right)$.

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

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

So:
 * $U \cap H \subseteq U \cap \operatorname{cl} \left({H}\right) = \left\{{s}\right\}$

There are two cases to address:

Either:
 * $s \in H$

or:
 * $s \notin H$


 * $U \cap H = \varnothing$

or:
 * $U \cap H = \left\{{s}\right\}$

depending on whether or not $s \in H$.

If $s \in H$ then:
 * $U \cap H = \left\{{s}\right\}$

and by definition $s$ is an isolated point of $H$.

If $s \notin H$, then: