The empty subset trivially satisfies the conditions of connectedness. Can you represent the empty subset as the disjoint union of two nonempty open sets?
Let . Then is also connected (I don't know why your book would not include singleton sets. They are also connected). Again, it cannot be represented by the disjoint union of two or more nonempty open sets.
Let with . Let . must be a finite set since is. So, let . Then for any , is an open subset of . Since every set containing a single point is an open subset of , if has more than one point, let and . By induction, you can show that is open and nonempty. And .