proof verification about the union of non-empty-intersection connected subspace is connected. I am a newbie on topological space and self-studying general topology when I read this pdf. I find some discrepancy on my proof and the author's proof.
Here is my problem:

If $\{A_i | i \in J\}$, is a collection of connected subspaces of a space X with $\bigcap A_i \neq \emptyset $, then $\bigcup A_i$ is connected.

My proof is:
Assume $M = \bigcup A_i $ is disconnected. There exist $U$ and $V$ in X such that $M \subset U \cap V$ and $M \cap U \cap V = \emptyset $.
Since $A_i$ is connected, if $A_i \cap U \cap V \neq \emptyset \implies U \text{ and } V \text{ are not separation for M }$, so $A_i \subset U$ or $A_i \subset V$. WLOG, $A_i \subset U$, there exists a $A_j$ which is a subset of $V$. Since  $\bigcap A_i \neq \emptyset $, then $A_i \cap A_j \neq \emptyset$ which turns out $U \cap V\neq \emptyset$
By $M \cap U \cap V = \emptyset $ and $M \subset U \cap V$, there are two cases: 
If $M \subset U$ and $M \cap V = \emptyset$ then $U$ and $V$ are not separation so it is contradict.
If $M \cap U \neq \emptyset$ and $M \cap  V \neq \emptyset$ then there exists $A_i$ and $A_j$ such that $A_i \cap A_j = \emptyset$ so it is contradict.
Is my proof right? Any help I will appreciate. ^-^
 A: Your proof is almost correct, in that it is correct (modulo a typo?) up to a point since you could declare victory earlier!
Proof 1 (Contradiction)
Suppose $M=\bigcup A_i$ is disconnected. Then there is a separation by open, nonempty sets of $M$, $M=U\cup V$. As you noted, if $A_i\cap U\ne\emptyset$, then by the connectedness condition, $A_i\subset U$. $V\ne\emptyset$, so, as you noted, $A_j\subset V$. But then $A_i\cap A_j\ne\emptyset$ as $\bigcap A_i\ne\emptyset$, as you noted. This a contradiction. Your proof can end here!
Proof 2 (Direct)
There is a useful characterization of connectedness by so-called "clopen" sets. If $X$ is a connected topological space, then the only clopen subsets are $X$ and $\emptyset$.
Let $U\subseteq M$ be open and closed (EDIT: and let $U$ contain a point of $\bigcap A_i$ !!!). Then $U\cap A_i$ is open and closed in the subspace topology of $A_i$, but $A_i$ is connected! Hence, $U\cap A_i=A_i$ and $A_i$ was arbitrary, hence, $\bigcup (U\cap A_i)=\bigcup A_i=M$, so $U=M$.
A: You have the righ idea, but your explanation is a little convoluted.
First, you need not assume $M$ is disconneced. You can start by assuming just
$M \subset U \cup V $ with $M \cap U \cap V = \emptyset  $ 
and show $M \subset U$ or $M \subset V .$
Once you get to where you say "WLOG, $A_i \subset U $", you could continue as follows (you do, but the exposition is confusing): the same reasoning applies to any other $A_j$ with $j \ne i$, but there could not be $j$ with $A_j \subset V $, otherwise we would have $ \bigcap_{t \in J} A_t \subset A_i \cap A_j \subset M \cap U \cap V = \emptyset $, which is a contadiction, therefore, since for one of the sets in $ \{A_t | t \in J\}$, namely $A_i$, we have $A_i \subset U $, then for all $j$ we have $A_j \subset U $, therefore $ M = \bigcup A_j \subset U$, which is what you wanted to show. 
