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

Your first part is O.K. Your second part is incomplete.
If $\partial A=\emptyset$ then $Cl(A) \setminus Int(A)=\emptyset$. It follows that
$Cl(A) \subseteq Int(A)$.
Since $Int(A) \subseteq A \subseteq Cl(A)$, we get
$$Int(A) = A = Cl(A).$$
Hence $A$ is clopen.
