My suggestion is to suppose that is not uniformly continuous. Then, there exists such that for every there exist such that but .
Try to construct a sequence that behaves badly. Keep in mind that the space is sequentially compact.
You first need to construct some kind of sequence out of the failure of uniform continuity. For example, for we can find points such that but . For , we can find such that but . Continuing the process, we can find such that but .
This gives you two sequences and such that and for every . Use sequential compactness here to relate them to the continuity of .
Note: This is more high-powered for than roninpro's suggestion. If it's gobbeldygook to you just ignore it.
It's trivial that for metric spaces sequential compactness and compactness are synonymous. So, let be arbitrary and cover with the open balls . Clearly then we have that is an open cover for . Then, since is compact we know that the open cover admits a Lebesgue number . Note then that if then we have that (this last part was pedantic, but just in case) and so by definition we have there is some such that and so and thus .