If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$.

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.

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 $X$.

Hint: Choose any $\varepsilon>0$ and $\delta>0$ such that $|x-y|<\delta\rightarrow |f(x)-f(y)|<\varepsilon$. Now, choose some $x\in (a,b)$. It is clear from definition that, for any $y\in (x-\delta,x+\delta)$ the value $|f(y)-f(x)|<\varepsilon$. By repeating this process on $y$, we can get that any value $z\in (y-\delta,y+\delta)$ has $|f(z)-f(x)|<2\varepsilon$ and hence, any value in $z\in (x-2\delta,x+2\delta)$ has $|f(z)-f(x)|<2\varepsilon$.

Proceed onto intervals of the form $(x-k\delta,x+k\delta)$ until you cover $(a,b)$.