differentiation I don't have any idea how to start solving this problem. Any help please?
Problem: 
Suppose that a differentiable function $f:\mathbb R \to \mathbb R$ and its derivative $f'$ have no common zeros. Prove that $f$ has only finitely many zeros in $[0,1]$.
 A: Suppose $(x_n)_{n\geq1}$ is a sequence of infinitely many distinct zeroes of $f$ in $[0,1]$. Since $[0,1]$ is compact, by eventually replacing this sequence with one of its subsequences, we can assume that there is a point $y\in[0,1]$ such that $\lim_{n\to\infty}x_n=y$. Since $f$ is continuous, this implies that $0=\lim_{n\to\infty}f(x_n)=f(y)$, so $f$ vanishes at $y$ also.
Now, since $x_n\to y$, we have $$f'(y)=\lim_{z\to y}\frac{f(z)-f(y)}{z-y}=\lim_{n\to\infty}\frac{f(x_n)-f(y)}{x_n-y}=0.$$ It follows that $y$ is a common zero of $f$ and $f'$.
A: A different argument.
Suppose $f$ and $f'$ have no common zeroes. If $x\in\mathbb R$ be a zero of $f$, then $f'(x)\neq0$ and $f$ is therefore injective in a small neighborhood $I=(x-\varepsilon,x+\varepsilon)$ of $x$: in particular, the only zero of $f$ in $I$ is $x$ itself. This shows that the set of zeroes of $f$ is discrete in $\mathbb R$ and, therefore, intersects every bounded closed interval in a finite set.
A: If the set of zeros of $f$ in $[0,1]$ is infinite, then it has a limit point in $[0,1]$.  Since $f$ is continuous, the limit point $x_0$ must be a zero of $f$.  There is a sequence $\{x_n\}_{n=1}^\infty$ of zeros of $f$ that approaches $x_0$.  So
$$
f'(x_0) = \lim_{n\to\infty} \frac{f(x_n)-f(x_0)}{x_n-x_0} = \lim_{n\to\infty} \frac{0-0}{x_n-x_0} = 0.
$$
So $x_0$ is a common zero of $f$ and $f'$.
A: Prove the contrapositive. Suppose $f$ is differentiable and it has infinitely many zeros in $[0,1]$. Now $[0,1]$ is compact, hence there exists a subsequence of zeros $x_{n_k}$ such that $f(x_{n_k}) = 0$. It also converges to a point where $f(x) = 0$ by continuity, and $x \in [0,1]$. This means that
$$
f'(x) = \lim_{k \to \infty} \frac{f(x_{n_k}) - f(x)}{x_{n_k} - x} = 0.
$$
Thus $x$ is a common zero of $f$ and $f'$.
