The interior of union of two boundary open sets is empty Show that if $U,V$ are two open sets of a topological space $X$ then  $$Int(\partial(U)\cup \partial(V))=\emptyset$$ 
I tried using that $\partial(U)=\overline{U}\cap \overline{U^c}$ but I can't solve the problem.
 A: Hints only.
Consider an open set $W\subseteq \partial U\cup\partial V.$ What can you say about $W$?
A: PART ONE. Lemma.Let $U$ be open. Then $\text{Int} (\partial U)=\emptyset.$ 
Proof: (1). $\partial U =\overline U \setminus U.$ 
Because $X\setminus U$ is closed so $\partial U=\overline U \cap \overline {X\setminus U}=\overline U\cap (X\setminus U)=\overline U \setminus U.$ 
(2). $\overline U \cap \text{Int} (\partial U)=\text{Int} (\partial U).$ 
Because $\text{Int}(\partial U)\subset \partial U\subset \overline U.$ 
(3). If $W$ is any open set that is  disjoint from $U$ then $\overline U \cap W=\emptyset.$
Because $X\setminus W$ is closed so $\overline U \cap W\subset \overline {X\setminus W}\cap W=(X\setminus W)\cap W=\emptyset.$ 
(4). Let $W=\text{Int} (\partial U) .$ Then we have $$U\cap W\subset U\cap (\overline U \setminus U) \subset U\cap ( X\setminus U) =\emptyset.$$ $$\text {Therefore }\quad  \emptyset =\overline U \cap W =\overline U \cap (\text{Int} (\partial U))=\text{Int} (\partial U). $$ End of proof of lemma.
PART TWO. Let $U$ and $V$ be open and let $W$ be any non-empty open set. The set $\partial U$ is closed. By Part One, $W^*=W\setminus \partial U$ is a non-empty open set (because $\text{Int}(\partial U)=\emptyset.$).  By Part One (again) $$W\setminus (\partial U\cup \partial V)=  W^* \setminus  \partial V \ne \emptyset$$ because $\text{Int} (\partial V)=\emptyset$ and $W^*$ is non-empty and open. 
So no non-empty open set $W$ is a subset of $\partial U \cup \partial V. $  Therefore $\text{Int}(\partial U\cup \partial V)=\emptyset.$
