$A$ is open $\iff$ $A\cap\partial A=\emptyset$ 
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.
 A: If you define the interior, boundary and exterior of a set in a topological space $X$ by


*

*$int(A)=\{x \in A : \text{ there exists an open set }U\text{ such that }x \in U \subset A\}$

*$ext(A)=\{x \notin A : \text{ there exists an open set }U\text{ such that }x \in U \subset X\setminus A\}$

*$\partial A= \{x \in X : \text{for any an open set }U\text{ such that }x \in U \text{ we have } U \cap A\neq \emptyset,\ U \cap (X \setminus A)\neq\emptyset \}$
It is easy to see that $X=int(A) \cup \partial A \cup ext(A)$ and the sets are disjoint. $A$ is open if and only if $A=int(A)$. Moreover for every $A$ we have $int(A) \subset A$. From the definitions above it is easy to see that $int(A) \cap \partial A=\emptyset$ and $A \subset int(A)\cup \partial A$.
Then if $A$ is open, $A=int(A)$ and $A \cap \partial A=\emptyset$.
If $A \cap \partial A=\emptyset$, then $A \subset int(A)$ and furthermore $A=int(A)$, which means that $A$ is open.
A: In your previous question, you had established the equality
$$A^{\circ} = A\setminus\partial A.$$
Since $A$ is open if and only if $A=A^{\circ}$, this gives:
$$\begin{align*}
A\text{ is open}&\Longleftrightarrow A=A^{\circ}\\
&\Longleftrightarrow A=A\setminus \partial A\\
&\Longleftrightarrow A\cap\partial A = \varnothing.
\end{align*}$$
Moral: Use previous results when possible!
