Union of an increasing sequence on connected sets Let $(A_i)$ be a family of strictly increasing connected sets, then I have to prove that $A=\bigcup_{i\in I}A_i$ is connected.
By contradiction, I assume that $A$ is not connected. Then there exist two open, non empty and disjoint sets $U$ and $V$ such that $ A= U\cup V,$
as $A_i$ is connected, we have $$\forall i\in I, [A_i\subset U~\text{or}~A_i\subset V]$$
How to prove that
 $$\forall i\in I, A_i\subset U~\text{or}~\forall i\in I, A_i\subset V$$
to get a contradiction?
Thank you.

@Ashish's comment I think makes it easier to prove. If you have $$ \forall i \in I, [A_i \subset U \text{ or } A_i \subset V]$$ and we know that $\exists i , A_i \subset U$ and $\exists j, A_j \subset V$ (because otherwise $U$ or $V$ would be empty), then your sets can't be increasing $A_i \not\subset A_j$ and $A_j \not\subset A_i$.
P.S. I assume that by increasing you mean that $\forall i, A_i \subset A_{i+1}$.
