$f$ would indeed be bounded.
$f$ is continuous on $(0,1)$; thus, if it were not bounded on $(0,1)$, there would exist $x_n\nearrow 1$ or $x_n\searrow 0$ with $f(x_n)\rightarrow\infty$ or $f(x_n)\rightarrow-\infty$. But, none of these options can happen due to the following facts: 1) uniformly continuous functions map Cauchy sequences to Cauchy sequences. 2) Cauchy sequences are bounded.
Fact 1) is easily proven: Let $\{x_n\}$ be Cauchy and $\epsilon>0$. Let $\delta$ be such that $|f(x)-f(y)|\lt\epsilon$ whenever $|x-y|<\delta$. Now choose $N$ so that $m,n> N$ implies $|x_n-x_m|<\delta$. We then have for any $n, m>N$ that $|f(x_n)-f(x_m)|<\epsilon$. Thus, $\{f(x_n)\}$ is Cauchy.
Fact 2) is easy to prove also, but I'll leave this to the interested reader.