Here is a proof from my notes that I could use some help with.

Let and be closed intervals. Suppose x is continuous. Show there are and in x , such that , for all x .

The hint that we are given is to divide the rectangle x into four subrectangles. Then, the supremum of over at least one of these subrectangles must be equal to the supremum of over x ...

I see that they are just applying the Extreme Value Theorem to rectangles but I'm not sure how to word the proof.