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**Pinkk** Here's my attempt:

Assume to the contrary that $\displaystyle f$ is a uniformly continuous function on a bounded set $\displaystyle S$ but $\displaystyle f$ is unbounded on $\displaystyle S$. Then there exists a sequence $\displaystyle (s_{n})\in S$ such that $\displaystyle |f(s_{n})|\ge n$ for all $\displaystyle n\in \mathbb{N}$. Since $\displaystyle (s_{n})$ is a bounded sequence, there exists a subsequence $\displaystyle (s_{n_{k}})$ that is a Cauchy sequence. Since $\displaystyle f$ is uniformly continuous and $\displaystyle (s_{n_{k}})$ is a Cauchy sequence in $\displaystyle S$, then $\displaystyle (f(s_{n_{k}}))$ is a Cauchy sequence. So $\displaystyle (f(s_{n_{k}}))$ converges to some real number. But $\displaystyle |f(s_{n_{k}})|\ge n_{k}$ for all $\displaystyle n_{k}\in \mathbb{N}$ as well, which is a contradiction. Q.E.D.

Is this correct? Thanks.