Show that $f$ cannot have infinitely many zeroes in $[0, 1]$. Let $f : \mathbb{R}\to \mathbb{R}$ be a differentiable function such that $f$ and its derivative have no common zero in the closed interval $[0, 1]$. Show that f cannot have infinitely many zeroes in $[0, 1]$.
 A: Since $[0,1]$ is compact, if $f$ had infinitely many zeroes there, they’d have to lie non-discretely, i.e. the zeroes of $f$ have a limit point in $[0,1]$. Because $f$ is continuous, that limit point would be a zero of $f$, too. Taking the derivative of $f$ at that limit point by choosing the zeros as approximating sequence for the difference quotients, one sees that the derivative of $f$ at that limit point would be zero as well. This is a contradiction. Therefore, $f$ cannot have infinitely many zeros in $[0,1]$.
A: Hints:
Suppose $\,\{x_n\}_{n\in\Bbb N}\,$ is an infinite sequence of zeros of $\,f\,$
1) Show there exists a subsequence $\,\{x_{n_k}\}\subset\{x_n\}\,$ s.t. that 
$$\lim_{k\to\infty}x_{n_k}=x_0\;,\;\;x_0\in [0,1]$$
2) Show $\,f(x_0)=0\,$
3) Show that between any two different zeros of $\,f\,$ there exists a zero of $\,f'\,$ (M.V.T.)
A: I assumed at first that the derivative is continuous which is clearly wrong, sry for that. 
Maybe you enjoy the example that it is not enough to have a function $f:[0,1]\to \mathbb{R}$ continuous in $[0,1]$ and continuous differentiable in $(0,1)$ (in fact it is $C^{\infty}$ in $(0,1)$
$$f(x)=\begin{cases}
x\cdot \sin\left(\frac{1}{x}\right) & x \neq 0\\
0& x=0
\end{cases}$$
