Given connected $(X,O)$ topological space , $\emptyset \ne A \subseteq X$ and $\partial(A)=\emptyset$: Show, $A = X$
Well, my idea:
$\emptyset$ = $\partial(A)$ = $\overline A \setminus A°$, which implies, that $\overline A = \emptyset$ or $\overline A = A°$. In case $\overline A = A°$, we can conclude, that $A=\overline A = A°$ because of $A° \subseteq A \subseteq \overline A$. That reveals that A is open and closed (since $A°$ is open and $\overline A$ is closed), which shows by definition ($X$ is connected <=> $A = X$, if $\emptyset \ne A \subseteq X$ is open and closed), that $A = X$ holds.
In case $\overline A = \emptyset$, we have that $A$ = $\emptyset$, because of $A \subseteq \overline A$. But we said, that $A \ne \emptyset$. That means, that this case is not going to happen anyway, so we are done.
Is my solution correct? Or is there something I missed/ have done wrong?