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?
 A: Suppose $\cup_\alpha$ A$_\alpha$ is disconnected. Then there exists two non-empty separeted sets E and F in X such that  $\cup_\alpha$ A$_\alpha$=E$\cup$F. Since each A$_\alpha$ is connected, either A$_\alpha$$\subset$E or A$_\alpha$$\subset$F for each $\alpha$. Without loss of generality let A$_1$$\subset$E. Since A$_1$$\cap$A$_2$$\neq$$\phi$, A$_2$$\subset$E. Proceeding as before we see that A$_\alpha$$\subset$E for each $\alpha$. Which contradicts that F is non-empty. Therefore $\cup_\alpha$ A$_\alpha$ must be connected.
A: We assume the union is disconnected, i.e. $\cup_\alpha^nA_\alpha=E\cup F$ where $E,F$ are open in the topology of the union, non-empty and disjoint. Now for any $n\in\Bbb N, E\cap A_n$ and $F\cap A_n$ are open in the subspace topology on $A_n$. If both these sets are non-empty, $A_n$ would be disconnected which is not the case, hence either $E\cap A_n$ or $F\cap A_n$ is empty, i.e. $A_n\subseteq E\oplus A_n\subseteq F\forall n\in\Bbb N$.
Since both $E,F$ are non-empty, there will exist $A_k,A_k+1$ such that $A_k\subseteq E$ and $A_{k+1}\subseteq F,k\ne j$. But since $E,F$ are also disjoint, this would mean $A_k,A_{k+1}$ are disjoint which contradicts the given condition.
A: Hint:

*

*Show that if $A, B$ are connected and $A \cap B \ne \emptyset$, then $A \cup B$ is connected.


*Use induction to prove your theorem.
Concerning 1: Pick $x \in A \cap B$. Then the connected component $C(x)$ of $x$ in the subspace $A \cup B$ contains the connected sets $A$ and $B$, thus $A \cup B \subset C(x) \subset A \cup B$. i.e. $C(x) = A \cup B$.
A: 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.
