The answer is in the question, let's prove it. Suppose we have $a\in {\bf R}$ with $f(a)>0$. Then $a\not\in {\bf Z}$ hence we have $n<a<n+1$ for some integer $n$. Choose $b\in [n,n+1]$ with $f(b)$ maximal, then $f(b)>0\implies b\in (n,n+1)\implies f'(b)=0 \implies f(b)=0$, contradiction. Hence $f\le 0$ everywhere and in the same way $f\ge 0$ everywhere. Hence $f\equiv 0$.

une blague

@above: let $A\subset {\bf R}$ be a nonempty subset such that for all $a,b\in A$ there exists a $c\in (a,b)$ with $c\in A$.That does not imply that $A$ is dense as the example $A=(-\infty,1/4)\cup (1/2,\infty)$ shows. So that alone is not enough to show that the zero locus of $f'$ is dense.

Correct, one can tack on the supposition that $A$ is additionally closed to make the implication true. Fortunately, this is indeed the case of the subspace in question. (Proof: It suffices to show that for arbitrary choice of point $p$, positive real $d$, and basic open $D(p,d)=\{x\in \mathbb R\text{ s.t. }|x-p|<d\}$, there exists an element of $A$ in $D(p,d)$. Suppose otherwise and observe that $A$ decomposes into disjoint closèds $^\dashv A:=A\cap (-\infty,p-d]$ and $A^\vdash:=A\cap (p+d,\infty]$. In particular, $\text{sup}(^\dashv A)\in ^\dashv A\subseteq (-\infty, p-d]$ and likewise $\text{inf}(A^\vdash)\in A^\vdash\subseteq [p+d,\infty)$. But this is a clear contradiction.)

This cannot be more simple than that? Single function with derivative to be $0$ is constant so $f = c$, $c \in R$. Hence from $f' = 0$ we need $f = 0$ we have $c = 0$ so only $f = 0$ is ok?

@rafa: there are functions that are differentiable but not $C^1$. Having IVP does not imply continuity. For an example https://math.stackexchange.com/questions/292275/discontinuous-derivative @ sticky : the second statement is not that for all $x$ we have $f'(x)=0$. Perhaps some extra brackets help: $\forall x (f'(x)=0\implies f(x)=0)$.

Claim : The function $f(x) = 0$ $\forall x \in [ n , n+1] $ for all $n \in \mathbb{N}$ Proof : $f'(n) = 0 \implies f(n) = 0 $ similarly $ f(n+1) = 0$ now as f is continuous it attains its extremas at $[n,n+1] $ if both the extremas are attained at end points we are done .Similarly if both are attained in the open interval then also we are done by fermats theorem and as maxima = minima = 0 .If not let $f(c) $ be an extreme value and the other extrema is attained at $n , n+1$ so if the function is not constant then $f (c) >0 $ (or $f(c) <0 $ but as by fermat $f'(c) = 0 \implies f(c) = 0 $ hence we are done and $f$ is constant and zero in [n,n+1] for all $n \in \mathbb{N}$ .Hence $f \equiv 0 $