Uniform continuity on (0,1) implies boundedness

Recall that if $f$ is uniformly continuous, then given $\epsilon>0$ we can find $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.

To show it is bounded, it doesn't really matter what $\epsilon$ is, so let it be some fixed constant. Let $N=\lfloor\frac{1}{\delta}\rfloor$ and take $$x_1=\delta,\ x_2=2\delta,\dots,\ x_n=n\delta,\dots, x_N=N\delta.$$ Notice every $y\in (0,1)$ satisfies $|y-x_i|<\delta$ for some $i$. Then $|f|$ will be bounded by $$\text{max}_{1\leq i\leq N} \{|f(x_i)|+\epsilon \}$$

which follows by applying the definition of Uniformly Continuous.

Hope that helps,

Here's one approach. You can use uniform continuity (with $\varepsilon=1$, say) to show that $f$ is bounded on $(0,\delta)$ and $(1-\delta,1)$. You probably already know the theorem that implies that $f$ is bounded on $[\delta,1-\delta]$.

If you don't know about boundedness of continuous functions on $[a,b]$, then what you can do here is cover $(0,1)$ with a finite number of tiny intervals where you know (using uniform continuity) that $f$ can't vary by more than $1$.

Uniformly continuous functions can also be extended to the closure, so an approach that would actually do more would be to show that $\lim_{x\to1}f(x)$ and $\lim_{x\to 0}f(x)$ exist, so that you may consider $f$ to be a restriction of a continuous function on $[0,1]$.