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'.