A subset $E⊆X$ is connected iff the only continuous functions $f:E→\{0,1\}$ are the constant ones. A continuous function $g:X→Y$ where $Y$ is Hausdorff is uniquely determined by a dense subset of $X$, and $\{0,1\}$ with the discrete metric is evidently Hausdorff. $E$ is dense in $\overline E$ and continuous functions $f:E→\{0,1\}$ are constant by assumption.
Another proof my proceed by contraposition. Suppose that $\overline E$ is disconencted by $U,V$. Then $E$ being dense has nonempty intersection with both $U,V$, so it is disconnected by $U'=E\cap U\;,\; V'=V\cap E$, for $U',V'$ are disjoint nonempty and relatively open in $E$, and $E=V'\cup U'$ by $\bar E=V\cup U$.
Suppose that $E$ is connected. Let $A,B\subseteq X$ be separated sets (that is, $\overline{A}\cap B=A\cap\overline{B}=\varnothing$) such that $\overline{E}=A\cup B$, and suppose that $A\neq\varnothing$. Let us prove that $B=\varnothing$.
Let $a\in A$. Since $A\cap \overline{B}=\varnothing$, there exists a neighborhood $U$ of $a$ such that $U\cap B=\varnothing$. Since $a\in\overline{E}$, then there exists some point $x\in E\cap U$, so $x\not\in B$, hence $x\in E\cap A$. Therefore, $E\cap A\neq\varnothing$.
Notice that $E=(A\cap E)\cup (B\cap E)$, and $A\cap E$ and $B\cap E$ are obviously separated. As $A\cap E\neq\varnothing$, from the previous paragraph, and $E$ is connected, then $B\cap E=\varnothing$.
(See PS below for an alternative end to the proof without the argument by contradiction)
Finally, suppose, in order to obtain a contradiction, that $B\neq\varnothing$, and take $b\in B$. By the same arguments as those used in the second paragraph above, switching $A$ and $B$ and $a$ by $b$, we would conclude that $B\cap E\neq\varnothing$, contradicting what we have just proved.
Therefore, $B=\varnothing$. This proves that $\overline{E}$ is connected.
PS: As $E\subseteq A\cup B$ and $E\cap B=\varnothing$, then $E\subseteq A$, so $\overline{E}\subseteq\overline{A}$. It follows that
$$B=B\cap\overline{E}\subseteq B\cap\overline{A}=\varnothing.$$
Best Answer
Since $U,V$ are nonempty and open in $\overline A$, and since $\overline A$ is the closure of $A$, each of $U,V$ has nonempty intersection with $A$. The sets $U'=U \cap A$, $V'=V \cap A$ therefore have the following properties: $A = U' \cup V'$; $U' \ne \emptyset$ and $V' \ne \emptyset$; $U',V'$ are open in $A$; $U' \cap V' = \emptyset$. It follows that $A$ is disconnected.