Let $A$ and $B$ be closed sets in the topological space $(X, \mathcal{T})$ with empty interiors. The task is to show that then $A\cup B$ has an empty interior as well.
What I've done is that I've assume by the way of contradiction that $\mathrm{int}(A\cup B) \neq \varnothing$. Then for some $x \in A\cup B$ there exists $U \in \mathcal{T}$ such that $x \in U \subset A \cup B$. Since $A$ and $B$ and $A\cup B$ are closed, they all contain their boundary. Thus, if $A$ and $B$ are disjoint we are done, as then either $U \subset A$ or $U\subset B$ contradicting the empty interiors of $A$ and $B$. So then, suppose that $A\cap B\neq \varnothing$. Then we must have that $U \subset A \cup B$, $U\not\subset A$ and $U\not\subset B$. And I don't know how to continue from here.
I'm struggling quite a bit to finish the claimed proof. Namely, I can't figure out the important connection between a supposed open neighborhood of a point $x \in A\cup B$ and the closedness of $A$ and $B$.
Best Answer
$U \subset A \cup B$ implies $U\setminus A \subset B$. Also, $U\setminus A=U\cap A^{c}$ is open. Since $B$ has no interior we conclude that $U\setminus A=\emptyset$. [Any point of this set would be an interior point of $B$]. But then $ x\in U \subset A$ so $x$ is an interior point of $A$, a contradiction.