Prove if $\partial (A)\cap \partial (B)=\emptyset,$ then $(A\cup B)^\circ=A^\circ\cup B^\circ$. I'm trying to prove the following result:

If $\partial (A)\cap \partial (B)=\emptyset,$ then  $(A\cup B)^\circ=A^\circ\cup B^\circ$.

The contention $\supseteq$ is straight forward. I'm having trouble proving the other way:
If $x\in(A\cup B)^\circ\implies\exists U\in\tau(x)$ s.t. $x\in U\subseteq A\cup B$. If either $U\in A\vee U\in B$, then the we easily get the result. Now, assume that $U\not\subseteq A,B$. The goal is to find an open set which is fully contained in either $A$ or $B$. Here I think I should use a property of boundaries and use that both are disjoint, but I don't see how to proceed.
I know the result of this post, but instead it is proven using that $\overline A\cap B=\emptyset=A\cap\overline B$. Does $\partial (A)\cap \partial (B)=\emptyset$ implies $\overline A\cap B=\emptyset=A\cap\overline B$?
 A: Generally, it is easy to find that $\left(A\cup B\right)^\circ\supset A^\circ\cup B^\circ$. We shall verify the inverse.
Suppose that $\left(A\cup B\right)^\circ\setminus\left( A^\circ\cup B^\circ\right)\neq \varnothing$, then choose $x\in \left(A\cup B\right)^\circ\setminus\left( A^\circ\cup B^\circ\right)$. For every neighbourhood $U$ of $x$, consider $U_0=U\cap \left(A\cup B\right)^\circ$, which is a neighbourhood of $x$ contained in $\left(A\cup B\right)^\circ$. If $U_0\cap A=\varnothing$, then $U_0\subset B$ and then $x\in B^\circ$, a contradiction and vice versa. Thus $U_0$ intersects both $A$ and $B$ and hence $U$ intersects both of them.
Thus every neighbourhood of $x$ intersects both $A$ and $B$ but $x$ is neither an interior point of $A$ nor $B$. Therefore every neighbourhood of $x$ interests both $A$ and $X\setminus A$ and both $B$ and $X\setminus B$, which implies that $x$ is in the boundary of both $A$ and $B$. Hence $\partial A\cap \partial B\neq\varnothing$.
A: Take $x\in (A\cup B)^\circ$. Then $x\in A^\circ$ or $x\in \partial A$. In the latter case, for any neighbourhood $U\subseteq A\cup B$ of $x$ we have $U\cap B\neq\emptyset$ (because $U\subseteq A\cup B$ and $U\not\subseteq A$). In particular, $x\in \overline{B}$, so that $x\in B^\circ$ from assumption that the boundaries have empty intersection.
