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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 $x\in \overline{H}?$

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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 $x \in \overline{H}$.

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  • $\begingroup$ This is right, of course, but it's not answering directly OP's question about his proof. $\endgroup$ – Daniel Jun 30 '16 at 21:33
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    $\begingroup$ I added a direct answer. $\endgroup$ – copper.hat Jun 30 '16 at 21:51
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For $x \bar {\bar H}:$ Observe that any open set $U$ containing $x$ is an open nbhd of every point of $U.$ Now there exists $y\in U\cap \bar H,$ and every nbhd of $y$ intersects $H$ (because $y\in \bar H$) so there exists $z\in U\cap H.$ That is $U\cap H\ne \emptyset$ for every open $U$ containing $x,$ so $x\in \bar H .$

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