Topological Closure of Subset is Subset of Topological Closure/Proof 2

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

Let $H \subseteq K$ and $K \subseteq S$.

Then:
 * $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$

where $\operatorname{cl}\left({H}\right)$ denotes the closure of $H$.

Proof
From the definition of closure:
 * $\operatorname{cl}\left({H}\right)$ is the union of $H$ and its limit points.

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

If $x \in H$ then $x \in K \implies x \in \operatorname{cl}\left({K}\right)$.

Otherwise $x$ is a limit point of $H$.

That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.

But as $y \in H$ it follows that $y \in K$.

So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.

This is the definition for a limit point of $K$.

Thus $x \in \operatorname{cl}\left({K}\right)$.