Showing that if the boundary of a set is connected, then the closure is connected. If $X$ is a connected metric space and $A\subseteq X$ has a connected boundary, then $\overline{A}$ is connected and $\overline{A^c}$ are both connected.
My attempted proof is as follows. As $\partial{A}=\overline{A}\cap\overline{A^c}$, then $\partial{A}\subseteq\overline{A}$. If $\overline{A}=C\cup D$, with $\overline{C}\cap D=\overline{D}\cap C=\emptyset$ with $C,D$ not empty.
Then $$\partial{A}=\partial{A}\cap\overline{A}=\partial{A}\cap(C\cup D)=(\partial{A}\cap C)\cup(\partial{A}\cap D)$$. But $\overline{(\partial{A}\cap C)}\cap(\partial{A}\cap D)\subseteq\overline{C}\cap D=\emptyset$ and similarly $\overline{(\partial{A}\cap d)}\cup(\partial{A}\cap C)=\emptyset$.
This means that $\partial{A}$ is disconnected, a contradiction, thus $\overline{A}$ is connected.
An analogous reasoning also shows that $\overline{A^c}$ is connected.
Is this correct, or am I missing something?
 A: I usually work with the definition that a set $S$ is connected if it cannot be written as a union of two disjoint nonempty sets which are both open (and closed) in $S$.
Assume that $C$ and $D$ are two disjoint nonempty sets which are both open (and closed) in $\overline{A}$. Since they are open in $\overline{A}$, there exist open sets $U,V \subseteq X$ such that $$C = U \cap \overline{A}, \quad D = V \cap \overline{A}.$$
We have that
$$C \cap \partial A = U \cap \partial A \quad\text{ and }\quad D \cap \partial A = V \cap \partial A$$
are disjoint sets which are both open in $\partial A$ and whose union is $\partial A$ so they cannot both be nonempty.
WLOG assume that $D \cap \partial A = \emptyset$. Then also $V \cap \partial A = \emptyset$ so we have
$$D = D \cap \overline{A} = V \cap (\operatorname{Int} A \cup \partial A) = (V \cap \operatorname{Int} A) \cup \underbrace{(V \cap \partial A)}_{=\emptyset} = V \cap \operatorname{Int} A$$
so we conclude that $D$ is open in $X$. Furthermore, $D$ is closed in $\overline{A}$ and $\overline{A}$ is closed in $X$ so $D$ is closed in $X$.
Hence $D$ and $X\setminus D$ (which contains $C\ne\emptyset$) are nonempty disjoint sets which are both open in $X$. This is a contradiction since $X$ is connected.
We conclude that $\overline{A}$ is connected.
A: I found the counterexample of an annulus with a radial line removed
