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**zebra2147** Well, I looked at all the links you gave me and they were all very interesting but I think I'm supposed to use the following lemma for this proof but extend it to rectangles...

Let $\displaystyle f:[0,1]\rightarrow \complex$ be continuous. Suppose $\displaystyle f$ is uniformly continuous on $\displaystyle I_{1}=[0,1/2]$, on $\displaystyle I_{2}=[1/4,3/4]$, and on $\displaystyle I_{3}=[1/2,1]$. Then $\displaystyle f$ is uniformly continuous on $\displaystyle [0,1]$.

That is why I previously stated my idea is to show that if $\displaystyle f$ is uniformly continuous on $\displaystyle [a,b]$ and is uniformly continuous on $\displaystyle [c,d]$ then f is uniformly continuous on $\displaystyle [a,b]$ x $\displaystyle [c,d]$??