Here's part 1.
This is a question from my tutorial that basically drags you through the proof. I'll post the question to let you have a go at it and if you can't answer it I'll start posting the solution! Note that the compliment of is the empty set. So if is both open and closed so is the empty set. Here's the question!
Let be a non-empty subset of and suppose that is both open and closed. Fix a point and define
Note that is non-empty (since is open).
Prove that is not bounded above. [Assume that it is and obtain contradictions both when sup , since is open, and when sup , since is closed.] Deduce that .
prove similarly that (here the contradictions arise by considering inf ).
Deduce that .
We have thus shown that the only subsets of that are both open and closed are and
the empty set.