Let $A_1, A_2, \dots $ connected subsets of $X$ s.t $A_n \cap A_{n+1} \ne \emptyset$ for all $n$. Show that the union of the sets $A_n$ is connected. 
Let $A_1, A_2, \dots $ be a sequence of connected subsets of a metric space $X$ such that $A_n \cap A_{n+1} \ne \emptyset$ for all $n$. Show that the union of the sets $A_n$ is connected.

So we want to show $\bigcup_\alpha ^n A_\alpha$ is connected. Assume the opposite so that $\bigcup_\alpha ^n A_\alpha$ is not connected, then $\bigcup_\alpha ^n A_\alpha = E \cup F$ for some sets $E$ and $F$. Pick $x \in \bigcup_\alpha ^n A_\alpha \implies x \in E$ or $x \in F$.
This is where I'm stuck, how can I proceed here?

Let $f: \bigcup_n A_n \to \{0,1\}$ be continuous. As each $A_n$ is continuous and $f\restriction_{A_n}$ is also continuous for each $n$, we have constants $c_n \in \{0,1\}$ so that $f\restriction_{A_n} \equiv c_n$ for each $n=1,2,3,\ldots$. Now I claim that $c_n =c_1$ for all $n$ by induction: ($n=1$ is trivial) and if it holds for $c_n$, then for $x \in A_n \cap A_{n+1}$ we have $c_n = f(x) = c_{n+1}$ so as $c_n=c_1$ it follows that $c_{n+1}=c_1$ too. So $f \equiv c_1$ and so $\bigcup_n A_n$ is connected.
