Show $A$ is open $\iff$ $A\cap\partial A=\emptyset$.
Attempt:
($\rightarrow)$ $A$ open $\implies A\cap\partial A= \emptyset$.
$x\in A$ open $\implies\exists\epsilon>0:B_{\epsilon}(x)\subseteq A$ $\implies$ $B_{\epsilon}(x)\cap A^c=\emptyset\implies$ $x\not\in\partial A$. Then $A\cap\partial A= \emptyset$.
($\leftarrow$) $A\cap\partial A= \emptyset$ $\implies A$ open.
$A\cap\partial A= \emptyset\implies$ for $x\in A$, $\exists\epsilon>0:B_{\epsilon}(x)\cap A^c=\emptyset\implies B_{\epsilon}(x)\subseteq A\implies A$ open.