If the derivative is zero on [a, b], then the function is constant - using Heine-Borel? I know the proof of the statement in the title using the mean value theorem but I was wondering if it can be proved using the Heine-Borel Lemma, which reads:
"Every open cover of a closed interval has a finite subcover." 
(Without compactness, simple as that). 
Thanks.
 A: Denote $I=[a,b]$ and let $f:I\to\mathbb R$ be a differentiable function whose differential vanishes identically.
Let $\epsilon>0$.
For any $x\in I$ we have $f'(x)=0$, so by definition of derivatives, there is $\delta_x>0$ such that $|f(x)-f(y)|<\epsilon|x-y|$ whenever $|y-x|<\delta_x$ and $y\in I$.
The open intervals $(x-\delta_x,x+\delta_x)\cap I$ cover $I$, and by Heine—Borel, finitely many of them are enough to cover it.
Let the midpoints be $a=x_1<x_2<\dots<x_n=b$ (we can always include the endpoints if we want).
For each $k$ pick a point $z_k\in (x_k,x_k+\delta_{x_k})\cap(x_{k+1}-\delta_{x_{k+1}},x_{k+1})\cap I$.
(Such a point exists since the $\delta$-intervals cover the interval $I$, as guaranteed by Heine—Borel.)
Now we have $a=x_1<z_1<x_2<z_2<x_3<\dots<x_{n-1}<z_{n-1}<x_n=b$ and each $z_k$ is within the "allowed $\delta$ range" from the points $x_k$ and $x_{k+1}$.
Take any $y\in I$. Suppose $y\in [x_m,z_m]$ for some $m$.
Then
$$
|f(a)-f(y)|
\leq
|f(x_1)-f(z_1)|+
|f(z_1)-f(x_2)|+
|f(x_2)-f(z_2)|+
|f(z_2)-f(x_3)|+
\dots+
|f(x_m)-f(y)|
\leq
\epsilon|x_1-z_1|+
\dots
\epsilon|x_m-y|
=
\epsilon(y-a)
\leq
\epsilon(b-a).
$$
The other option is that $y\in[z_m,x_{m+1}]$ for some $m$, and you get the same estimate.
Since $|f(a)-f(y)|\leq\epsilon(b-a)$ for any $y\in I$ and $\epsilon>0$, you must have $f(a)=f(y)$ for all $y\in I$.
Therefore $f$ is constant.
