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

Let $\displaystyle I$ and $\displaystyle J$ be closed intervals. Suppose $\displaystyle f:I$ x $\displaystyle J\rightarrow R$ is continuous. Show there are$\displaystyle (x_{min},y_{min}) $and $\displaystyle (x_{max},y_{max})$ in $\displaystyle I$ x $\displaystyle J$, such that $\displaystyle f(x_{min},y_{min})\leq f(x,y)\leq f(x_{max},y_{max})$, for all $\displaystyle (x,y)\in I$ x $\displaystyle J$.

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

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