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If $\operatorname{Bd}A \cap \operatorname{Bd}B= \emptyset$, show that $\operatorname{int}(A)\cup\operatorname{int}(B)=\operatorname{int}(A\cup B)$

One of inclusions hold without the assumption

$\operatorname{int}(A)\subseteq A$

$\operatorname{int}(B)\subseteq B$

So $\operatorname{int}(A)\cup \operatorname{int}(B)\subseteq A\cup B$

Then taking "int" for both sides we get $\operatorname{int}(A)\cup\operatorname{int}(B)\subseteq\operatorname{int}(A\cup B)$ (since $\operatorname{int}(A)\cup\operatorname{int}(B)$ is open, then its interior is itself)

Now to prove the inverse inclusion we need to use the assumption. I tried to prove it by contradiction (i.e if $x\notin\operatorname{int}(A)\cup\operatorname{int}(B)$ then want to show $x \notin\operatorname{int}(A\cup B)$)

But I face some problems while trying to reach my goal.

Can any one help me?

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    Use that for each subset $C$,$X =\operatorname{int}(C) \cup \operatorname{bd}(C) \cup \operatorname{ext}(C)$, where $\operatorname{ext}(C) = \operatorname{int}(X \setminus C)$ and this union is disjoint. – 2017-02-11

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Suppose $x\in\operatorname{int}(A\cup B)$ but $x\notin \operatorname{int}(A)\cup\operatorname{int}(B);$ I claim that $x\in\operatorname{Bd}(A)\cap\operatorname{Bd}(B).$ I will show that $x\in\operatorname{Bd}(A);$ the proof of $x\in\operatorname{Bd}(B)$ is similar. Since $x\notin\operatorname{int}(A),$ it will suffice to show that $x\in\operatorname{Cl}(A).$ Let $U$ be any neighborhood of $x.$ Then $V=U\cap\operatorname{int}(A\cup B)$ is a neighborhood of $x,$ and $V\subseteq A\cup B,$ but $V\not\subseteq B$ (since $x\notin\operatorname{int}(B)),$ so $V\cap A\ne\emptyset$ and therefore $U\cap A\ne\emptyset.$