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. 

Let $x_n\to 1$, then $\{x_n\}_{n\in\mathbb N}$ is a Cauchy sequence, and so is $\{f(x_n)\}_{n\in\mathbb N}$, due to uniform continuity. This implies that $f$ extends continuously to $x=1$ and similarly to $x=0$. Hence, $f$ extends to a continuous function on a compact set $[0,1]$. Therefore, $f$ is bounded.
Note. In general, if $D$ is dense in $X$, and $f:D\to\mathbb R$ is uniformly continuous, then $f$ extends uniquely to a continuous function on $X$.
