Verification of proof that if $S$ is connected then the closure of $S$ is connected. I attempted to prove this theorem by contrapositive. Suppose $\overline{S}$ 
is disconnected. Then $\overline{S}=A\cup B$ such that $\overline{A}\cap B=A\cap \overline{B}=\varnothing$ and $A$ and $B$ are both nonempty. It follows that since $S\subseteq \overline{S}$, $S\subseteq A\cup B$. If $S$ is closed then the theorem follows trivially, so we will consider $S$ not closed. If $S\subset A\cup B$, then $S\subset A$, $S\subset B$, or $S$ contains some elements from $A$ and 
some from $B$. Suppose $S\subset A$, then $\overline{S}\subset \overline{A}$, and therefore there must be a $p\in\overline{S}$ that is also in $\overline{A}$ and $B$ since $B$ is nonempty and $\overline{S}=A\cup B$. But this contradicts $\overline{A}\cap B=\varnothing$. The same arguments holds for $S\subset B$. 
Therefore it must be the case that $S$ contains elements of $A$ and $B$. Define the following sets, 
$C=\{p : p\in S\cap A\}$
$D=\{q :q\in S\cap B\}$. 
Then $S=C\cup D$. Consier $\overline{C}\cap D$. If $p\in \overline{C}\cap D$, then $p\in \overline{A}\cap B$. Another contradiction. The same argument holds for $C\cap \overline{D}$. Therefore $S=C\cup D$ with 
$\overline{C}\cap D=C\cap \overline{D}=\varnothing$. Thus, by contrapositive, if $S$ is connected then $\overline{S}$ is connected. 
 A: Instead of with separated subsets, work with non-trivial clopen sets (it's equivalent): suppose $C$ is a non-empty clopen subset of $\overline{S}$; being open it intersects $S$ non-emptily, so $C \cap S$ is non-empty clopen in $S$. As $S$ is connected, $C \cap S= S$, or equivalently $S \subseteq C$. It follows that $\overline{S} \subseteq \overline{C}=C$ and so $C=\overline{S}$ and $\overline{S}$ is connected.
Yet another alternative: suppose $f: \overline{S} \to 2=(\{0,1\}, \tau_{\text{discr}})$ is continuous; we need to show it's constant. $f\restriction_S: S \to 2$ must be constant by connectedness of $S$ and as $S$ is dense in $\overline{S}$ and the space $2$ is Hausdorff, we get that $f$ is constant on $S$ as it coincides with a constant map on a dense subset. QED. 
Or if you insist on separated sets: Suppose that $\overline{S}= A \cup B$, a separation. 
Then $S = (S \cap A) \cup (S \cap B)$ and the intersection sets are still separated, of course. So, as $S$ is connected, one of them is empty, say $S \cap B=\emptyset$ and the other equals $S$ so $S \cap A=S$.
But then $S \subseteq A$, hence $\overline{S} \subseteq \overline{A} \subseteq \overline{S}\setminus B \subseteq A$, which shows that $B$ must be empty and we're done.
A: The OP is using the following characterization of a disconnected space $X$,

$X$ cannot be written as the union of two non-empty separated sets (sets for which each is disjoint from the other's closure).

(see this wikilink).
Of course a $\text{<NOT Disconnected>}$ space $X$ is called a connected space.
In the next section we directly answer the OP's question using this characterization as the definition.

Proposition: If $\overline{S}$ is disconnected then $S$ is disconnected.
Proof
Suppose $\overline{S}$ 
is disconnected. Then $\overline{S}=A\cup B$ such that $\overline{A}\cap B=A\cap \overline{B}=\emptyset$ and $A$ and $B$ are both nonempty. Since $A \cap B \subset \overline{A} \cap B = \emptyset$, clearly the subsets $A$ and $B$
constitute a 2-block partition of $\overline{S}$.
If $S \subset A$, then $\overline{S} \subset \overline{A}$ (see this). But then since $\overline{A}\cap B=\emptyset$,  $\,\overline{S}\cap B=\emptyset$, contradicting the premise that $\{A,B\}$ is a 2-block partition of $\overline{S}$. So $S \cap B \ne \emptyset$,
Using a symmetric argument, $S \cap A \ne \emptyset$.
So $\{S \cap A, S \cap B\}$ is a 2-block partition of $S$,
$$\tag 1 S = [S \cap A] \;  \bigcup \; [S \cap B]$$
To show that $S$ is discounted we'll use the theory found in this math.stackexchange post
$\quad$Closure of a subset of a subspace of a topological space
So, denoting the closure of $S \cap A$ in $S$ by $\overline{S \cap A}^{(S)}$, we have
$$\tag 2 [\overline{S \cap A}^{(S)}] \cap [S \cap B] = [S \cap \overline{S \cap A}] \cap  [S \cap B] $$
We have 
$\quad \overline{S \cap A} \subset \overline{S} \cap \overline{A} \quad$ (the closure of an intersection of sets is always
$\quad \quad \quad \quad \quad \quad \quad \quad \;$ a subset of the intersection of the closures of the sets)
and since  $\overline{A}\cap B =\emptyset$, we must conclude that
$\quad [\overline{S \cap A}^{(S)}]  \; \bigcap \; [S \cap B] = \emptyset$
Symmetric logic allows us to write
$\quad [S \cap A] \; \bigcap \;[\overline{S \cap B}^{(S)}] = \emptyset$
We have shown that $S$ is disconnected. $\quad \blacksquare$
