$Cl(Int(Cl(A \cup B))) = Cl(Int(Cl(A))) \cup Cl(Int(Cl(B)))$ Let $A,B$ be two sets in topological space
Prove or disprove:
$$\operatorname{Cl}(\operatorname{Int}(\operatorname{Cl}(A \cup B))) = \operatorname{Cl}(\operatorname{Int}(\operatorname{Cl}(A))) \cup \operatorname{Cl}(\operatorname{Int}(\operatorname{Cl}(B)))$$
where $\operatorname{Cl}$ is closure, and $\operatorname{Int}$ is interior.
I have tried to find a counterexample, but it seems to be true, I have no idea..
 A: Note that since $\DeclareMathOperator{\uswCLevi}{Cl}\DeclareMathOperator{\tuzINazt}{Int}\DeclareMathOperator{\umzBDsym}{Bd}\uswCLevi ( A \cup B ) = \uswCLevi ( A ) \cup \uswCLevi ( B )$ it suffices to show for closed $F , E \subseteq X$ that $$\uswCLevi ( \tuzINazt ( F \cup E ) ) = \uswCLevi ( \tuzINazt ( F ) ) \cup \uswCLevi ( \tuzINazt ( E ) ).$$
 For $A \subseteq X$ let $\umzBDsym A$ denote the boundary of $A$, that is, $\umzBDsym ( A ) = \uswCLevi ( A ) \setminus \tuzINazt ( A )$. We need a few simple facts.
Fact 1. For $A , B \subseteq X$, $\tuzINazt ( A \cup B ) \subseteq \tuzINazt ( A ) \cup \tuzINazt ( B ) \cup \umzBDsym ( A ) \cup \umzBDsym ( B ) $.


*

*proof. Clearly $\tuzINazt ( A \cup B ) \subseteq A \cup B \subseteq \uswCLevi ( A \cup B ) = \uswCLevi ( A ) \cup \uswCLevi ( B )$, and use the fact that for $D \subseteq X$, $\uswCLevi ( D ) = \tuzINazt ( D ) \cup \umzBDsym ( D )$.


Fact 2. If $F \subseteq X$ is closed, then $\tuzINazt ( \umzBDsym ( F ) ) = \emptyset$.


*

*proof. As $S$ is closed, then by definition $\umzBDsym ( F ) = \uswCLevi ( F ) \setminus \tuzINazt ( F ) = F \setminus \tuzINazt ( F )$. In particular $\tuzINazt ( \umzBDsym ( F ) ) \subseteq F$, and since $\tuzINazt ( \umzBDsym ( F ) )$ is open we have $\tuzINazt ( \umzBDsym ( F ) ) \subseteq \tuzINazt ( F )$. But as $\umzBDsym ( F )$ is disjoint from $\tuzINazt ( F )$ it must be that $\tuzINazt ( \umzBDsym ( F ) ) = \emptyset$.



We now move on to the proof.
For the simple inclusion, note that as $F \subseteq F \cup E$, then $\tuzINazt ( F ) \subseteq \tuzINazt ( F \cup E )$, and so $\uswCLevi ( \tuzINazt ( F ) ) \subseteq \uswCLevi ( \tuzINazt ( F ) )$. Similarly $\uswCLevi ( \tuzINazt ( E ) ) \subseteq \uswCLevi ( \tuzINazt ( F \cup E ) )$.
For the opposite inclusion, let $x \in \uswCLevi ( \tuzINazt ( F \cup E ) )$, and let $U$ be any open neighborhood of $x$. We must show that $U \cap ( \tuzINazt ( F ) \cup \tuzINazt ( E ) ) \neq \emptyset$. As $x \in \uswCLevi ( \tuzINazt ( F \cup E ) )$ using the two Facts above we have
$$\begin{align}
\emptyset &\subsetneqq U \cap \uswCLevi ( F \cup E ) \\
&\subseteq U \cap ( \tuzINazt ( F ) \cup \tuzINazt ( E ) \cup \umzBDsym ( F ) \cup \umzBDsym ( E ) ) \\
& = U \cap ( \tuzINazt ( F ) \cup \tuzINazt ( E ) \cup \emptyset \cup \emptyset ) \\
&= U \cap ( \tuzINazt ( F ) \cup \tuzINazt ( E ) )
\end{align}$$
as desired.
