Zero derivative implies constant function (No MVT, Rolle's Theorem, etc.)

Let $a < x < y < b$. We will prove $f(y)=f(x)$ by proving $|f(y)-f(x)| < \epsilon(y-x)$ for every $\epsilon>0$. So fix $\epsilon > 0$. By the hypothesis on vanishing of derivatives, for every $t \in [x,y]$ there is $\delta_t > 0$ such that for $|z-t|<\delta_t$, $|f(z)-f(t)|<\epsilon|z-t|$.

Now here is a minor little trick: The intervals $(t-\delta_t/2,t+\delta_t/2)$ (of radius $\delta_t/2$, rather than $\delta_t$) cover $[x,y]$. For now say an interval of the form $(t-\delta_t/2,t+\delta_t/2)$ is a narrow interval and one of the form $(t-\delta_t,t+\delta_t)$ is a wide interval. By compactness there is a finite subcover by narrow intervals, centered at say $t_1,\dotsc,t_s$, with $x \leq t_1 < t_2 < \dotsc < t_s \leq y$. We can reduce to an irredundant cover, i.e., so that no narrow interval centered at $t_i$ is contained in the narrow interval centered at $t_j$ for $j \neq i$. This implies that for each $i$, the narrow intervals centered at $t_i$ and $t_{i+1}$ overlap: otherwise, at least one point in the gap between them must be covered by the interval centered at some $t_j$ with $j<i$ or $j>i+1$; and either $t_i$ is redundant, or $t_{i+1}$ is redundant. Next, this implies that for each $i$, $|t_{i+1}-t_i| < \max(\delta_{t_{i+1}},\delta_{t_i})$. Indeed, if a point $u$, $t_i < u < t_{i+1}$, lies in the intersection of the narrow intervals, then $$ t_{i+1}-t_i = t_{i+1}-u+u-t_i < \delta_{t_{i+1}}/2 + \delta_{t_i}/2 \leq 2 \max(\delta_{t_{i+1}}/2,\delta_{t_i}/2). $$ It follows that for each $i$, $t_{i+1}$ lies in the wide interval centered at $t_i$ or $t_i$ lies in the wide interval centered at $t_{i+1}$ (or both). In particular, for each $i$, $|f(t_{i+1})-f(t_i)| < \epsilon |t_{i+1}-t_i|$.

We claim that the same relation holds between $x$ and $t_1$, and between $y$ and $t_s$. Indeed, $x$ lies in the narrow interval centered at $t_1$, hence also in the wide interval. So $|f(t_1)-f(x)| < \epsilon |t_1-x|$. Similarly for $y$ and $t_s$.

Putting it all together, we have: $$ \begin{split} |f(y)-f(x)| &\leq |f(y)-f(t_s)| + |f(t_s)-f(t_{s-1})| + \dotsb + |f(t_1)-f(x)| \\ &< \epsilon|y-t_s| + \epsilon|t_s-t_{s-1}| + \dotsb + \epsilon|t_1-x| \\ &= \epsilon(y-t_s) + \epsilon(t_s-t_{s-1}) + \dotsb + \epsilon(t_1-x) \\ &= \epsilon(y-x), \end{split} $$ as claimed.

Notes: (1) Thanks for several very helpful comments about simplifications, gaps in the proof, etc! (2) I wonder if an application of Cousin's theorem might simplify the argument.

Note that since $f$ is differentiable on $(a,b),$ $f$ is continuous on $(a,b).$

Suppose $a<c<d<b.$ We'll show $f(c)-f(d)=0,$ which implies $f$ is constant. Let $\epsilon>0.$ Define

$$E=\{x\in [c,d]:|f(x)-f(c)|\le \epsilon(x-c)\}.$$

Then $c\in E,$ so $E$ is nonempty. Let $s= \sup E.$ Then there is a sequence $x_n$ in $E$ such that $x_n\uparrow s.$ The continuity of $f$ then shows $s\in E.$

Suppose $s<d.$ Because $f'(s) = 0,$ there exists $x\in (s,d)$ such that $|f(x)-f(s)|\le \epsilon(x-s).$ We then have

$$|f(x)-f(c)|\le |f(x)-f(s)| + |f(s)-f(c)| \le \epsilon(x-s)+\epsilon(s-c) = \epsilon(x-c).$$

Thus $x\in E$ and $x> s,$ contradiction. Therefore $s=d,$ and since we know $s\in E,$ we have $d\in E.$ That means $|f(d)-f(c)|\le \epsilon(d-c).$ Since $\epsilon$ was arbitrary, we have $f(d)=f(c)$ as desired.