Union of connected space 
If $\{A_\alpha, \alpha \in I\}$ is a collection of conected spaces and $\cap A_\alpha \neq \emptyset$ then $\cup A_\alpha$ is connected.

My proof: If $\cup A_\alpha$ is not connected then we can find $V,W\neq\emptyset$ such that $V\cup W = \cup A_\alpha$ and $V\cap W = \emptyset$. Suppose $\cup A_\alpha$ is not connected. Since $\cap A_\alpha \neq \emptyset$ then $\exists x\in \cap A_\alpha$. Because $V\cup W = \cup A_\alpha$ and $V\cap W = \emptyset$ $x\in V$ or $x\in W$, but if $x\in V$ then $x\notin W$ therefore $\exists \beta \in I$ such that $x\notin A_\beta$, likewise if $x\in W$. We reach a contradiction, thus $\cup A_\alpha$ is connected

Suppose $\cup_{\alpha} A_{\alpha}$ is not connected and let $V$ and $W$ be two nonempty disjoint sets so that $V\cup W = \cup_{\alpha} A_{\alpha}$.  Let $x\in \cap_{\alpha} A_{\alpha}$.  Then $x \in V$ or $x \in W$ but not both.  Wolog let $x \in V$.
Let $y \in W$, then $y \in A_i$ for some set $A_i$.  And $x = A_i$ because $x$ is in all $A_{\alpha}$.  So $W\cap A_i$ is not empty as it contains $y$ and $V \cap A_i$ is not empty as it contains $x$.  So $A_i = (W\cap A_i) \cup(V\cap A_i)$ two disjoint non-empty sets.
Now $A_i$ is connected so either $\overline{ (W\cap A_i)}\cap  (V\cap A_i)$ is not empty or $(W\cap A_i)\cap  \overline{(V\cap A_i)}$
Let $p$ be a point in the non-empty intersection.  If $p \not \in W$ then $p \in V$ but $p$ is in $\overline{ (W\cap A_i)}$ so it is a limit point of $ (W\cap A_i)$ therefore a limit point of W and so $\overline W \cap V \ne \emptyset$.
Otherwise $p \in W$ but  $p \not \in V$ and the same argument shows $\overline{V} \cap W \ne \emptyset$.
So $\cup_{\alpha} A_{\alpha}$ can not be partitioned into two disjoint nonempty sets so that the closure of neither intersect with the other.  So the union is connected.
