 Solution

Use the separation definition in (i). If $A={\bigcup }_i{A}_i$ had a separation $A=U\cup V$, the intersections with each connected $A_i$ show that $A_i$ lies wholly in one side. Choose one nonempty $A_{i_0}$; every other $A_i$ intersects it and must lie on that same side. Thus the other side is empty, a contradiction. If the union is empty it is connected by definition. Therefore $\boxed{\bigcup _i{A}_i\text{ is connected}}$.