Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous. Prove the following statements:
$(a)$ For any $\epsilon>0$, there exists a piecewise constant function $s:[a, b] \rightarrow \mathbb{R}$ such that $\|f-s\|_{\infty}<\epsilon$
$(b)$ If $f$ is uniformly continuous and strictly monotone, inverse function $f^{-1}$ is also uniformly continuous.
Regarding $(a)$, I found that $f$ is uniformly continuous, and we should use the epsilon-delta definition of it; however, after several attempts, I still could not come up with a solid example for it.
About $(b)$, I could only prove the existence of inverse function and how it transforms the open set in range to the open set in domain.
(1) Fix the $\epsilon >0$. Then we know there is a $\delta > 0$ such that if $x,y$ are $<\delta$ apart then $f(x), f(y)$ are $<\epsilon$ apart. Define $s$ on chunks of length $2\delta$:
i.e, for $x \in [a, a+ 2\delta)$, we have a constant function $s_1(x) = f(a + \delta)$. Similarly, for the next $x \in [a + 2\delta, a + 4\delta)$, set a constant function $s_2(x) = f(a+ 3\delta)$. Can you see why this works? Take $s_1$ for instance: if $x \in [a, a+2 \delta)$, then we can easily see that $|x - (a + \delta)| <\delta$ which means that $|f(x) - f(a+ \delta)| = |f(x) - s_1(x)| < \epsilon$.
Splice together these constant functions (of which there are finitely many of course) to gain one piecewise constant function $s$.
(2) Since the domain is a bounded interval and the function is strictly monotone, then $dom(f^{-1}) = ran(f) = [f(a), f(b)]$ (or $[f(b), f(a)]$) is a closed boundeded interval as well. You already found out that $f^{-1}$ is continuous. Then simply use the fact that a continuous function on a closed bounded interval is uniformly continuous.