Show that if $A$ is a connected subset of a metric space $X$, then $\overline{A}$ is connected.
My approach: If $A$ is a connected subset of metric space $X$, then $\emptyset$ and $A$ are open and closed.
If $A=A^{\circ}$, then $A$ is open and if $A=\overline{A}$, then $A$ is closed. So,
$$A^{\circ}=\overline{A}^{\circ}\subset \overline{A}$$ $$\overline{A}=A\subset\overline{A}=A^{\circ}=\overline{A}^{\circ}\implies \overline{A}\subset\overline{A}^{\circ}$$
Then $\overline{A}=\overline{A}^{\circ}=A^{\circ}$, then $\overline{A}$ is open. And clearly, $\overline{A}$ is closed. So $\overline{A}$ is closed. This is correct?? Thanks!
You have the definition of connected wrong. $A$ is connected iff its only clopen subsets (when it is considered a subspace) are itself and $\emptyset$.
Suppose $\bar A$ is not connected. Then there is nonempty, clopen (in $\bar A$) $B$ with $B\subsetneq\bar A$.
Then consider $C=B\cap A$.
Claim: $C\subsetneq A$ is nonempty and clopen in $A$.
$C$ is nonempty because $B\subset\bar A$ is open.
$C$ is clopen in $A$ because $B=U\cap \bar A$, where $U$ is clopen in $X$ (thus $C=U\cap A$).