Does this variant on Rolle's theorem have a name?

This variant is stated as exercice V.1.12 in the book Problems and Theorems in Analysis (Tome 2) by Polya and Szego, and is there an anonymous early consequence of Rolle's theorem. So I don't think it has a well-known name.

You can state it as $$ Z(f) \le Z(f')+1, $$ where $Z$ gives the number of zeros of a function.

Note. Pick up a copy of Problems and Theorems in Analysis if you can. It's an accessible read and the exercices start easy in each section.

You reach a wrong conclusion. A function can be continuously differentiable and have infinitely many zeros in an open interval, without being constant in any subinterval.

Consider $$ f(x)=\begin{cases} 0 & \text{if $x=0$} \\[6px] x^3\sin\dfrac{1}{x} & \text{if $x\ne0$} \end{cases} $$ Then $f$ is everywhere differentiable and $$ f'(x)=\begin{cases} 0 & \text{if $x=0$} \\[6px] 3x^2\sin\dfrac{1}{x}-x\cos\dfrac{1}{x} & \text{if $x\ne0$} \end{cases} $$ which is continuous. On the other hand, there is no interval where $f$ is constant.

If the set $Z(f)$ of zeros of $f$ is finite, say $Z(f)=\{x_0,x_1,\dots,x_n\}$, we can define $\varphi\colon Z(f)\setminus\{x_0\}\to Z(f')$ by assigning to $x_i$ an arbitrarily chosen zero of $f'$ in the interval $(x_{i-1},x_i)$, whose existence is guaranteed by Rolle’s theorem. The choice can be made canonical by choosing the leftmost point of absolute maximum in $(x_{i-1},x_i)$ if $f$ assumes positive values in the interval, the leftmost point of absolute minimum otherwise (note that $f$ cannot be constant over the subinterval).

If $Z(f)$ is infinite, the situation is hairier. In the case it is countable, the set contains either an increasing or a decreasing sequence. Rolle’s theorem can be easily applied in the same way as before.

What about the uncountable case?