Closure of a connected set is connected Let $(X,d)$ be  a metric space and let $E \subseteq X$ connected. I want to show that $\overline E$ is connected.
How can I prove this in a nice way ?
 A: Following Daniel Fischer's suggestion:
Suppose that $E$ is a connected subspace of $X$ and $E \subseteq K \subseteq \overline E$.
Consider $K$ as a subspace of $X$. Then $E$ is a dense, connected subspace of $K$.
Let $C$ and $D$ be open sets in $K$ that separate $K$.
Then $C$ and $D$ are nonempty, and thus each must contain an element of $E$, so $C\cap E$ and $D \cap E$ separate $E$.
A: Suppose that $X$ is a space and $E\subset X$ is a connected subset.
Let $A$ be a clopen subset of $\overline E$ such that $A\cap\overline E\not=\emptyset$.
Since $A$ is open in $\overline E$, $A\cap E\not=\emptyset$.
Note that $A\cap E$ is nonempty and clopen in $E$.
Since $E$ is connected, $E=A\cap E\subset A$.
Since $A$ is closed in $\overline E$, $\overline E\subset A$.
So $A=\overline E$, showing $\overline E$ to be connected.
A: 
PROP Let $E\subseteq X$ be connected. Then $\overline E$ is connected.

P Let $f:\overline E\to\{0,1\}$ be continuous. Since $E$ is connected, $f\mid_E$ is constant. But $E$ is dense in $\overline E$, so $f$ is constant, since continuous functions are entirely determined on a dense subset of their domain whenever the codomain is Hausdorff. And $\{0,1\}$ with the discrete topology is metrizable, hence Hausdorff, the claim follows.
ADD There is a more general claim. Let $(X,\mathscr T)$ be a topological space. If $E$ is connected and $K$ is such that $E\subseteq  K\subseteq \overline E$, then $K$ is connected.
P Consider $K$ as a subspace of $X$. Then $E$ is dense in the space $K$. Let $f:K\to\{0,1\}$ be continuous. Then $f\mid_E$ is constant. It follows $f$ is constant, so $K$ is connected.
Here is the proof of 

PROP Let $X$, $Y$ be topological spaces, $Y$ Hausdorff. Suppose $D$ is dense in $X$ and $f,g:X\to Y$ are continuous. If $f$ and $g$ agree on $D$, then $f=g$.

P By contradiction. Thus, assume there exists $x\in X\smallsetminus D$ such that $f(x)\neq g(x)$. Then there exist open nbhds of $N_1$ of $f(x)$ and $N_2$ of $g(x)$ with $N_1\cap N_2=\varnothing$. By continuity, $M_1=f^{-1}(N_1)$ and $g^{-1}(N_2)=M_2$ are open. Then so is $M=M_1\cap M_2\neq \varnothing$ since $x\in M$. Thus, there exists $y\in M\cap D$, and $f(y)=g(y)$. But this is impossible, since this gives $f(y)\in f(M)\subset ff^{-1}(N_1)\subset N_1$ and $g(y)\in g(M)\subset gg^{-1}(N_2)\subseteq N_2$.
A: Suppose $E$ is connected and its closure $\bar E$ is disconnected then there exists sets $P$ and $Q$ such that $P \cup \bar Q = \bar P \cup Q = \phi$ and $ P \cup Q = \bar E$.
Now let $P \cap E = M$ and $Q \cap E = N$
$M$ and $N$ have the property of $M \cup \bar N = \bar M \cup N = \phi$ as they are subset of $P$ and $Q$.
Also $M \cup N = E$ which shows that $E$ is disconnected and hence we have a contradiction.
