Here's what I need to prove:
if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on $S$.
I plan on using this theorem: If $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence.
Here's my attempt:
Assume that $f$ is not bounded on $S$, then for every $M>0$, there is $x \in S$ such that $|f(x)|>M$. It follows that there is a sequence $(s_n)$ in $S$ such that $|f(s_n)| > n$ for every $n \in \mathbb{N}$. Now, since $(s_n)$ is bounded (because $S$ is bounded), by the Bolzano Weierstrass Theorem, there is an convergent subsequence $(s_{n_{k}})$. Since, $(s_{n_{k}})$ is convergent, it is Cauchy as well. However, $|f(s_{n_{k}})|$ does not converge to a finite value and hence is not Cauchy. This contradicts that theorem.
Is this proof correct or do I go wrong somewhere?