If $E$ is closed, then $\partial \partial E = \partial E$. 
Prove that if $E$ is closed, then $\partial \partial E = \partial E$.

I found proofs for this when in a metric space, but I would like to show this in general topological space without the usage of neighborhoods.
I’ve managed to get the problem down to the part where I need to show that $\operatorname{int}(\partial E) = \emptyset$.
So what I have is the following. $$\partial \partial E  = \operatorname{cl} (\partial E) \setminus \operatorname{int}(\partial E).$$
Since $\partial E$ is a closed set $\operatorname{cl(\partial E)}  = \partial E$ and thus $$\partial \partial E = \partial E \setminus \operatorname{int}(\partial E).$$
If $\operatorname{int}(\partial E)  = \emptyset$ that would complete the proof, but apparently this is not neccessarily true?
 A: It is true.
Let $U = \operatorname{int}( \partial E )$. Then $U \subseteq \partial E$ and since $E$ is closed, $\partial E \subseteq E$ so $U \subseteq E$. It follows that $U \subseteq \operatorname{int}( E )$ so $U \cap \partial E = \varnothing$. On the other hand $U \subseteq \partial E$, hence $U = \varnothing$.
A: The interior of the boundary of a closed set is the empty set. You can find this proof on Wikipedia, but I'll rewrite it here for completeness.
Let ${\displaystyle U}$ be an open set such that ${\displaystyle U\subseteq \partial E}$. Then ${\displaystyle U\subseteq E}$ (because ${\displaystyle \partial E\subseteq E}$ as $E$ is closed), so that ${\displaystyle U\subseteq \operatorname {int}E}$ (because by definition, ${\displaystyle \operatorname {int} E}$ is the largest open subset contained in ${\displaystyle E}$).
But ${\displaystyle U\subseteq \partial E=E\setminus \operatorname {int}E}$ implies that ${\displaystyle U\cap \operatorname {int} E=\varnothing .}$ Thus ${\displaystyle U}$ is simultaneously a subset of ${\displaystyle \operatorname {int} E}$ and disjoint from ${\displaystyle \operatorname {int} E,}$ which is only possible if ${\displaystyle U=\varnothing .}$
A: Let $\overline {A}$ be the closure and $A^o$ the interior of $A$. Then $\:\partial (A)=\overline {A}-A^o$.
If $E$ is closed, then
$$
\partial (E)^o=(E-E^o)^o=E^o\cap((E^o)^c)^o\subset E^o\cap(E^o)^c=\varnothing
$$
So $\:\partial (\partial (E))=\overline{\partial (E)}-\partial (E)^o=\partial (E)$.
A: For any set $A$,
$$ \partial A = \overline{A}\cap\overline{(X\setminus A)}=\partial (X\setminus A)$$
Since $X\setminus \overline{B}= \operatorname{Int}(X\setminus B)$, or equivalently $\operatorname{Int}(C)=X\setminus\overline{(X\setminus C)}$ we have that
$$\begin{align}
\partial A&= \overline{A}\setminus\operatorname{Int}(A)\\
\end{align}$$
If $A=\overline{A}$, then
$$\begin{align}
\partial A&= A\setminus\operatorname{Int}(A)
\end{align}$$
and so,
\begin{align}
X\setminus\partial A &= (X\setminus A)\cup \operatorname{Int}(A)
\end{align}
Recall that for any two sets $U, V\subset X$, $\overline{U\cup V}=\overline{U}\cup\overline{V}$. Hence
$$\begin{align}\partial A \cap \overline{(X\setminus \partial A)}&=\big(A\cap\overline{(X\setminus A)}\big)\cap\overline{\big((X\setminus A)\cup \operatorname{Int}(A)\big)}\\
&=\big(A\cap\overline{(X\setminus A)}\big)\cup\big(A\cap\overline{(X\setminus A)}\cap\overline{\operatorname{Int}(A)}\big)\\
&=A\cap\overline{X\setminus A}=\partial A
\end{align}$$
since $A\cap\overline{(X\setminus A)}\cap\overline{\operatorname{Int}(A)}\subset \overline{A}\cap\overline{(X\setminus A)}=\partial A$.
