Prove that if f : (a,b)-->R is is uniformly continuous function in (a,b) so f is bounded function.
Assume it's not bounded then pick a (unbounded) sequence $\displaystyle (x_n) \subset (a,b)$ such that $\displaystyle f(x_n)<f(x_{n+1})$ for all $\displaystyle n$ then since this sequence is bounded we get a Cauchy subsequence $\displaystyle (x_{n_k})$ by Bolzano-Weierstrass. Since $\displaystyle f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $\displaystyle f(x_{n_k})$ by construction?.
I believe that now it is easier to understand it.
If $\displaystyle f$ is unif. cont and $\displaystyle \left(x_{n_k}\right)$ is Cauchy, then $\displaystyle f\left(x_{n_k}\right)$ is Cauchy too.
Let $\displaystyle \epsilon>0$. Since $\displaystyle f$ is unif cont, there exists $\displaystyle \delta>0$ such as $\displaystyle |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon.$ Since $\displaystyle \left(x_{n_k}\right)$ is Cauchy, there existe $\displaystyle K$ such as $\displaystyle |x_{n_i}-x_{n_j}|<\delta$ for all $\displaystyle i,j\geq K$.
So you know a Cauchy sequence is bounded, and Abu-Khalil proved that if $\displaystyle f$ is a unif. cont. function and $\displaystyle x_n$ is a Cauchy seq. then $\displaystyle f(x_n)$ is also a Cauchy seq. but by how we constructed them $\displaystyle f(x_n) \rightarrow \infty$ (or $\displaystyle -\infty$ in which case invert all inequalities in my first post) clearly contradicting the fact that $\displaystyle f$ is unif. cont. so we conclude that our initial hypothesis is false ie. $\displaystyle f$ is bounded.