$\{A_n\}$ is a sequence of connected subspace of $X$ s.t. $A_n \cap A_{n+1} \not= \emptyset$ for all $n$. Show that $\cup A_n$ is connected I know there are several answers to this questions on M.S., but I found them difficult to understand. I also found some answers to this question from some professors' solutions but they are also difficult to understand. I used induction to prove it on my homework, but it's obviously wrong since induction just proves finite union. Could anyone prove it without defining a function?

A solution I found is posted below. Does proving $B_n$ is connected by induction on $n$ really prove the infinite union? I still think it just proves finite union.

 A: Well I don't know what theorem this solution is referring to, but here's how we can complete the proof:
If $\cup_n B_n$ were disconnected, there would exist nonempty open disjoint subsets $C_1,C_2$ with $C_1\cup C_2=\cup_nB_n$.  Since $B_1$ is connected, it lies in either $C_1$ or $C_2$. Assume $B_1\subset C_1$.  Then since $B_n\subset B_{n+1}$ you can show by induction that $B_n\subset C_1$ for all $n$.  Thus $C_2=\varnothing$, a contradiction.
A: If The union is not connected then there are two open disjoint sets $O_1$ and $ O_2$  such that $A_n \subset O_1$ for some $n\in S\subset  \mathbb{N}$ and  $A_n \subset O_2$ otherwise.
Without loss of generality assume that $A_1 \subset O_1.$ 
We claim that for some n, $A_n\subset O_1$ and $A_{n+1}\subset O_2.$ 
Otherwise, every $A_n$ is a subset of $O_1$ which contradicts our assumption.
Thus we have $A_n \cap A_{n+1} = \phi,$ which is not possible.    
A: The OP might find a direct proof easier to understand.

The OP has a chain
$\tag 1 B_1 \subset B_2 \subset B_3 \dots \subset B_{n} \subset \dots$
of nonempty connected sets the union of which is the same as the union of the $A_n$.
Let $U$ and $V$ be any two disjoint open sets of $B = \bigcup_{\,n \ge 1} B_n$ such that $B = U \cup V$. All the $B_n$ are nonempty connected subspaces, so they are all contained in either $U$ or $V$. Since $B_1$ is contained in every subsequent $B_n$ in the chain, if $B_1$ is contained in, say $U$, then $B_n \subset U$ for all $n \in \mathbb N$. But then $U$ must be equal to $B$ and $V$ is empty. Similarly, if $B_1 \subset V$, then $U = \emptyset$.
We have demonstrated that $B$ is connected.
