Problem with closure of a topological closure 
Prove closure of a topological closure is equal to the topological closure.

If we denote the closure of a set $H$ of a topological space $X$ with $\overline H$ then using the definition of a closure implies $\overline{(\overline H)}$ is the union of $\overline H$ and its limit points, hence $\overline H⊆\overline{(\overline H)}$
on the other hand let $x∈\overline{(\overline H)}$,then from an equivalent condition for being in the closure of a set it follows that every open neighborhood $U$ of $x$ has a point in common with $\overline H$, denoted $y$
e.g. $$\color{green}{U∩\overline H≠∅}$$
since $y$ is in $U∩\overline H$ hence it's also in $\overline H$, so again every open neighborhood of $y$ intersects $H$ at a point denoted $z$, but how do we know $z$ is also in $U$?
I think since $y$ is in $U$ so we can find some neighborhoods of $y$ such that these neighborhoods intersect $H$ at a point (namely $z$), but as I said I think we can find them, but it's not possible to say every arbitrary open neighborhood of $y$ have a point in common with $H$
I've seen a proof of this theorem at this site, but I still have not been convinced. 
 A: Let $x \in \overline{\overline{A}}$, where $A \subseteq X$. We want to show $x \in \overline{A}$. That $\overline{A} \subseteq \overline{\overline{A}}$ is indeed already trivial, so we still need the reverse inclusion.
I use the criterion you mention for $x \in \overline{B}$ for any set $B$:
$$x \in \overline{B} \iff \forall O \subseteq X \text{ open }: (x \in O) \to (O \cap B \neq \emptyset)$$
So let $O$ be in with $x \in O$. We know (by applying the criterion to $x,O$ and $\overline{\overline{A}}$) that $O \cap \overline{A} \neq \emptyset$, so pick $y \in \overline{A} \cap O$. Now we apply the same criterion to $y$ and $O$ for $\overline{A}$, we have that $O \cap A \neq \emptyset$ and so (as $O$ was an arbitrary open neighbourhood of $x$), $x \in \overline{A}$, as required.
A: Let $y\in \overline{\overline{H}}$. Then, for each open neighborhood $U$ of $y$, we have $U\cap\overline{H}\ne\emptyset$. Let $z\in U\cap\overline{H}$. Then $z\in\overline{H}$ and, being $U$ an open neighborhood of $z$, we conclude that $U\cap H\ne\emptyset$.
Just recall that every neighborhood contains an open neighborhood, so it's not restrictive to assume $U$ to be open, and that an open set is a neighborhood of each of its points, so $U$ is a neighborhood of $y$ as well as of $z$.
