Union of connected sets also connected? I got this question for homework.. 
Let $\{E_{\alpha}\}$ be a family of connected subsets of a metric space $X$ such that any
two of them have a non-empty intersection: $E_{\alpha} \cap E_{\beta}\ne \emptyset$. Prove that the union $\cup_{\alpha}E_{\alpha}$ is connected.
I'm relatively new to the concept of connected sets. I know that a set is connected iff it is the union of 2 separated sets and that one of the two sets are empty.
I don't understand how the fact that some of these sets share elements have anything to do with the union of all these sets... 
Please help!
Thanks
 A: We will argue by contradiction. Assume that $\cup_{\alpha}E_{\alpha}$ is not connected. Then there exist $U,V$ non-empty, open and disjoint such that $\cup_{\alpha}E_{\alpha}=U\cup V.$ Now define $f:U\cup V\to \{0,1\}$ by $$f(x)=\begin{cases}0 ,& x\in U, \\ 1,& x\in V.\end{cases}$$ It is clear that $f$ is continuous. Let $x_0\in U.$ There exist $\alpha$ such that $x\in E_{\alpha}.$ Since $f$ is continuous and $E_{\alpha}$ is connected then $f(E_{\alpha})$ must be connected. That is $f(E_{\alpha})=\{0\}.$ Now, for any $\beta$ we have $E_{\alpha} \cap E_{\beta}\ne \emptyset.$ So we get that $f(E_{\beta})=\{0\}.$ That is, $f$ must be constant. But this contradicts our assumption that $V\ne \emptyset.$ So we are done. 
Note that the assumption $E_{\alpha} \cap E_{\beta}\ne \emptyset$ is crucial. In other case think of $(0,1)$ and $(1,2).$ Is it their union connected?
A: In my answer a separation of $Y\subseteq X$ is by definition a pair $\{A,B\}$ of non-empty sets with $A\cup B=Y$ and $\bar{A}\cap B=\emptyset=A\cap\bar{B}$. Subset $Y$ is connected if no separation of $Y$ exists.

Suppose that $\{A,B\}$ is a separation of the union. Then $A\cap B=\varnothing$ and : $$\forall\alpha [E_{\alpha}\subseteq A\vee E_{\alpha}\subseteq B]$$
This because the negation of it for some $\alpha_0$ would lead to the conclusion that this $E_{\alpha_0}$ is not connected: pair $\{A\cap E_{\alpha_0},B\cap E_{\alpha_0}\}$ would be a separation of $E_{\alpha_0}$. 
Then the condition that $\beta\neq\alpha\implies E_{\alpha}\cap E_{\beta}\neq\varnothing$ can be used to show that:$$\forall\alpha\; E_{\alpha}\subseteq A\vee\forall\alpha\; E_{\alpha}\subseteq B$$
However then $B=\varnothing$ or $A=\varnothing$, contradicting that $\{A,B\}$ is a separation.
We conclude that no separation exists, i.e. that the union is connected.
