If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$. So I know that since $f$ is uniformly continuous on $(a,b)$, then for every 
$\varepsilon > 0$, there exists $\delta > 0$ such that for all $x$ and $y$ in $(a,b)$, 
if $\,\lvert x - y\rvert < \delta$, then $\,\lvert\,f(x) - f(y)\rvert < \varepsilon$. 
I also know that I need to show that there are some numbers $M$, $N$, so that 
$M ≤ f(x) ≤ N$, for all $x$ and $y$ in $(a,b)$. 
It makes logical sense to me because it means that for every x and y that you pick, the function values can only jump by epsilon, so there would be no way for the function suddenly approach infinity at the endpoints. However, I don't know really how to make a rigorous argument. I was thinking that maybe I could do something by contradiction, like saying that $f(y)>N$, but $f(x)<N$, and then somehow showing that this jump is now greater than 
$\varepsilon$, but I'm not sure that would really work since we only have an arbitrary function with arbitrary bounds. 

Indirect proof: Suppose $f$ is not bounded above; choose $x_0$; then there exists $x_1$ such that $f(x_0)+1< f(x_1)$ (otherwise $f(x_0)+1$ would be a bound); now you may continue: there exists $x_2$ such that $f(x_2)> f(x_1)+1$, and so on. By construction, for $i\ne j$, $\vert f(x_i)-f(x_j)\vert >1$. By assumption there exists $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert < 1$. Now you just have to argue that among the elements of the sequence $(x_i)$ constructed at the beginning there are $x_i,x_j$, with $i\ne j$ such that $\vert x_i-x_j\vert<\delta$ (why?) which leads to a contradiction.
