Assume there is no $x ∈ R$ such that $f(x) = f'(x) = 0$. Show that $$S =\{x: 0≤x≤1,f(x)=0\}$$ is finite.

We prove this by contradiction. Assume $S$ is not finite. Since $S \subset [0,1]$, $S$ is bounded and hence it must have a converging subsequence. So there exists $x_n \in S$ such that 
\[f(x_n)=0,\]
and $\lim_{n\rightarrow \infty} x_n=x_0 \in [0,1]$ with $x_n \neq x_0$. Since $f$ is continuous, 
\[f(x_0)=\lim_{n\rightarrow \infty} f(x_n)= \lim_{n\rightarrow \infty} 0=0.\]
Thus 
\[f'(x_0)=\lim_{x \rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{n \rightarrow \infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}\]
\[=\lim_{n \rightarrow \infty}\frac{0-0}{x_n-x_0}=0.\]
Hence $f(x_0)=f'(x_0)=0$, $x_0 \in [0,1]$ which is a contradiction. So $S$ must be finite.