So the only possible point of discontinuity is . To check this we first notice three things: is a limit point of , that , and are left and right hand continuous respectively. So
Now remembering that is the unique point in such that we can conclude that , thus is continuous on .
Now is here where I am not sure what level of math you are at. Tell me if this is beyond the scope of your course
Theorem: If is continous and is compact, then so is
Proof: Let be an open covering of . Using the fact that the inverse image of any open set in must be open in we can see that is an opening cover of . But since is compact it follows that . Using the identity we can concude that thus is compact.
So knowing by the Heine-Borel Theorem that any closed set is compact we can conclude that is compact, thus closed and bounded, and due to its boundedness we know that the supremum and infimum are included.
So back to our actual problem since we have that . So since is both bounded and closed it is compact, and since is continuous it follows that so is , thus , which is what was asked to be proven.
I'll come back later for number two.
Note: was gotten by being continuous and being connected...in case you are in analysis.