Let $A \subset X$ be connected. Show that $\overline{A}$ is also connected. 
Let $A \subset X$ be connected. Show that $\overline{A}$ is also connected.

Suppose that $\overline{A}$ is disconnected, then $\overline{A} = G \cup H$ for open sets $G$ and $H$ that satisfy $G \cap H = \emptyset$. Then since $A \subseteq \overline{A}$ and $A$ is connected we must have that $A \cap G = \emptyset$ or $A \cap H = \emptyset$ i.e $A \subseteq G$ or $A \subseteq H$. Without loss of generality assume that $A \cap G = \emptyset$, then $$A\subseteq H \implies \overline{A} \subseteq \overline{H}$$ but then $\overline{A} = \overline{H}$ which results in a contradiction since $\overline{A} = \overline{G} \cup \overline{H}$.
Is the proof correct? I think I could have chosen either one of the intersections to contain $A$ and it would still work?
 A: Suppose that $f: \overline{A} \to \{0,1\}$ is continuous where the codomain is discrete. Then $f\restriction_A: A \to \{0,1\}$ is also continuous and so constant with value $i \in \{0,1\}$. By continuity, $f[\overline{A}] \subseteq \overline{f[A]} = \overline{\{i\}} = \{i\}$ and so $f \equiv i$ as well. It follows that $\overline{A}$ is connected.
In your proof you have to specify in what space $G$ and $H$ are open and in what space are you taking the closures? As it stands your proof is unclear. The above approach avoids such issues altogether.
Yours can be fixed, though: suppose $\overline{A}=G \cup H$ where $G,H$ are disjoint and open in $\overline{A}$ so that $G=G' \cap \overline{A}$ and $H=H' \cap \overline{A}$ for open subsets $G',H'$ of $X$. As $A = (G' \cap A) \cup (H' \cap A)$ as $A \subseteq \overline{A}$ and both sets are open in $A$ and disjoint on $A$, we have WLOG $A= G' \cap A$, say (and $H'\cap A=\emptyset$), so $A \subseteq G'$ and $A \subseteq X\setminus H'$; the last set being closed in $X$ gives $\overline{A} \subseteq X\setminus H'$ (closures in $X$) so $$H'=\overline{A} \cap H \subseteq \overline{A} \cap H' = \emptyset$$ so $\overline{A} = G$ and we're done.
