The closure of a connected set in a topological space is connected This problem is from Rudin. I am trying to Prove that the closure of a connected set is always connected. Here is my proof.
Let $E$ be a connected set in a space $X$. Suppose to the contrary that the closure of $E$, $\overline{E}$ is not connected. Then there exist two sets $A$ and $B$ such that $\overline{E}=A \cup B$ and $\overline{A}\cap B= \emptyset=A\cap\overline{B}$. $E$ being connected, we know that $A\cup B \neq E$ so there exist $p \in \overline{E} \backslash E$. We also know that $\overline{E}=E\cup E'$ (where $E'$ is the set of the limit points of $E$) so $p$ must be a limit point of $E$ (but not in $E$). Taking the set of all such $p$ we obtain the set $E''=\{p|p\in E', p\not \in E\}$. We find then that $E'' \subset A$ or $E'' \subset B$. Say $E'' \subset A$. Then $E \subset B$ and we see that $A\cap\overline{B}\neq \emptyset$ which is a contradiction. So $\overline{E}$ must be connected.
Can anyone help me critique this proof? I feel uneasy about it but I don't know exactly what is wrong with it.
Thank-you
 A: 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
$$B=B\cap\overline{E}\subseteq B\cap\overline{A}=\varnothing.$$
A: Yes.
For any topological space $X$, let $E$ be the connected set . To prove that $E$ closure is connected, suppose $\mathrm{Cl}(E)$ is disconnected. Then $\exists$ at least two non-empty open sets say $H$ and $K$ in $\mathrm{Cl}(E)$ such that $\mathrm{Cl}(E)= H\cup K$.
Since $H$ and $K$ are open in $\mathrm{Cl}(E)$ and $E$ contained in $\mathrm{Cl}(E)$, so $H\cap E$ and $K \cap E$ are non-empty disjoint open sets in $E$ such that :

$$E= (H \cap E) \cup (K \cap E)$$

Which gives us that $E$ is disconnected, which is a contradiction to the fact that $E$ is connected.
Thus, our supposition is wrong.
Hence, $\mathrm{Cl}(E)$ is connected.
A: I believe this can be made even more concise: Suppose $\overline{E}=A\cup B$ for disjoint, nonempty, and open $A,B$.
$E$ connected and $E=(A\cap E)\cup (B\cap E)$, so wlog, $A\cap E=E$.
Then $B$ is an open set containing a limit point of $E$, and so it must intersect $E\subseteq A$ nontrivially - contradiction, as $A\cap B=\emptyset$.
A: There is only one  part which  might not have been explained in detail  **

$ E''\subset A$ or $E''\subset B$

**
$A$ and B are separation of $\bar{E}$ implies $A\cap E$ and $B\cap E$ is a separation of $E$(trivial to proof). 
$\implies$ $\overline{(A\cap E)}\cap(B\cap E)=\emptyset$ $\implies (\bar{A}\cap\bar{E})\cap B \cap E=\emptyset (\because \bar{X}\cap \bar{Y}\subset \overline{X\cap Y}  )\implies \bar{A}\cap B \cap \bar{E}=\emptyset \implies A\cap B\cap\bar{ E}=\emptyset$  
(I kept using the fact: $C\cap D=\emptyset $ and $C'\subset C$ then $C'\cap D=\emptyset$)
$A\cap B\cap \bar{E}=\emptyset$  says if $x\in \bar{E}$ and also $x\in A$ then $x\not\in B$ (Similarly,  $x\in \bar{E}$ and also $x\in B$ then $x\not\in A$.
Therefore, $E''\subset A$ or $E''\subset B$
It is important to keep using the equivalent definitions of connectedness:
A topological space $X$ is disconnected if
Definition 1:  there are two non-empty open sets $A$ and $B$ such that $X=A\cup B$ and $A\cap B=\emptyset$
Definition 2: there are two subsets $A$ and $B$ such that $X=A\cup B,$  $\bar{A}\cap B=\emptyset$ and $A\cap \bar{B}=\emptyset$
A: Theorem: $(X, \tau) $ be a topological space. $A\subset X$ connected, then $\overline{A}$ is also connected.
Proof:

A topological space $(Y, \tau') $ is connected iff every continuous map from $(Y, \tau') $ to a two-point discrete space is constant.

Let $f:\overline{A}\to (\{0, 1\},\tau_{\text{discrete}})$ be a continuous map.
Since $A$ is connected, the map $f|_{A} : A\to (\{0, 1\},\tau_{\text{discrete}})$ is constant.
WLOG suppose $f(A) =\{0\}$
Then $f=0$ on $A$ implies $f=0 $ on $\overline{A}$ (see here)
Implies $f(\overline{A})=\{0\}$
Hence the map $f:\overline{A}\to (\{0, 1\},\tau_{\text{discrete}})$ is $\text{constant}$. $\color{red}{\boxed{\text{done}}}$
