My first attempt:
Since uniform continuity implies continuity, $f((a,b))$ is an open interval in $\mathbb{R}$, which is bounded, thus $f$ is bounded.
Is this correct?
I was not convinced myself enough, so tried different approach (second attempt):
Fix $\epsilon>0$, then for all $x,y\in(a,b)$, there exists $\delta>0$ such that $$|x-y|\leq\delta\implies |f(x)-f(y)|<\epsilon.$$
And \begin{align} |f(c)| &= |f(c)-f(c+\delta)+f(c+\delta)| \\ &\leq |f(c)-f(c+\delta)|+|f(c+\delta)| \\ &<\epsilon+|f(c+\delta)| \\ &... \\ &<n\epsilon + |f(c+n\delta)| \end{align} where $c\in(a,b)$, $n$ satisfies $c+(n+1)\delta\geq b$.
I tried to bound $|f(c)|$ by a combination of $a,b,\epsilon$ and $\delta$, but it seemed not to work out.
For your first try is incorrect, for example $f(x)=\frac{1}{x}$ on the interval $(0,1)$.
As for your second try, it's the right direction and can be made to work. First, we let $\varepsilon=1$. By uniform continuity, we have $\delta>0$ s.t $|x-y|<\delta$ implies $|f(x)-f(y)|<1$. Now we look at the closed interval $[a+\frac{\delta}{2},b-\frac{\delta}{2}]\subset (a,b)$ (we can assume $\delta$ is small enough s.t there is inclusion).
$f$ is continuous and hence bounded by some number $M$ on the closed interval we chose. Now let $x\in (a,b)$. If $x\in [a+\frac{\delta}{2},b-\frac{\delta}{2}]$ then $|f(x)|\leq M<1+M$. Otherwise if $x\in (a,b)\setminus[a+\frac{\delta}{2},b-\frac{\delta}{2}]$ there exists some $y\in [a+\frac{\delta}{2},b-\frac{\delta}{2}]$ s.t $|x-y|<\delta$. For this specific $y$ we have: $$|f(x)|\leq |f(x)-f(y)|+|f(y)|\leq 1+M$$ So $f$ is bounded by $1+M$.