Easy exercise with boundary could I have a confirm or a suggestion about this little exercise?

$\partial A=\emptyset$ if and only if $A$ is open and closed.

Sol.:
If $A$ is "clopen", then $Int(A)=A$ and $Cl(A)=A$, so $\partial A=A \setminus A=\emptyset $.
If $\partial A=\emptyset$, then $Cl(A) \setminus Int(A)=\emptyset$. So doesn't exist a $x \in Cl(A) \cap Int(A)$. Then $Cl(A)=Int(A)$, and this is possibile only if $A$ is clopen

I find easier this approach:
The boundary of $A$ is not empty $\iff$ there exists some $x\in\ Cl(A)\cap Cl(A^c)$ $\iff$ there exists some $x$ that has no neighbourhood contained in $A$ or $A^c$ $\iff$ $A$ is not open or $A^c$ is not open $\iff$ $A$ is not clopen.
