Q:

Problem in understanding the proof of closure of topological closure is closure.

Well, I was following the proof of Closure of Topological Closure equals Closure; here is the proof:

It follows directly from Set is Subset of its Topological Closure that:
$$\overline{H} \subseteq \overline{\left(\overline{H}\right)}$$
$\Box$
Let $x \in \overline{\left(\overline{H}\right)}$
Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x\in U$ contains some $y\in \overline{H}$
If we consider $U$ as an open set containing $y,$ it follows that:
$$U \cap H \ne \varnothing$$
Hence, $x\in \overline{H}$.
$\blacksquare$

I could all but the last line; how did they conclude lastly that $x\in \overline{H}?$

A:

It follows from the definition. The closure of $A$ is the smallest closed set containing $A$.
By definition we have $\overline{H} \subset \overline{\overline{H}}$.
Since $\overline{H}$ is a closed set containing $\overline{H}$, we have
$\overline{\overline{H}} \subset \overline{H}$, and so
$\overline{\overline{H}} = \overline{H}$
Addendum:
A point $x$ is in the closure of $A$ iff for all open $U$ that contain $x$, then $U \cap A \neq \emptyset$.
If $x \in \overline{\overline{H}}$ and $U$ is an open set containing $X$,
then $U \cap {\overline{H}} \neq \emptyset$. Since $U$ is open, and contains
a point in $\overline{H}$, then it must intersect $H$ and so $U \cap H \neq \emptyset$. Since this is true for all open $U$ containing $x$, we see that
$x \in \overline{H}$.

