$|f''(x)|\leq 1$, $\exists\ x_1\neq x_2, \in [0,1]$ such that $f(x_1)=f(x_2)=0$. Show that $|f(x)|\leq 1, \forall\ x\in [0,1].$ 
$|f''(x)|\leq 1$, $\exists\ x_1\neq x_2, \in [0,1]$ such that $f(x_1)=f(x_2)=0$. Show that $|f(x)|\leq 1, \forall\ x\in [0,1].$

If $x_1=0, x_2=1$, then Taylor expansion would help. But here we only know that $x_1\neq x_2$! What this helps?

Since $f(x_1)=f(x_2)$ then by Rolle's theorem $f'(c) = 0$ for some $c$ between $x_1$ and $x_2$.
If $f''$ is continuous then we have $ \left| f'(x) \right|=\left| \int_{c}^{x} f''(t)dt \right|\leq \left|x-c\right|$ for all $x \in [0,1]$.
Similarly ,  $ \left| f(x) \right|=\left| \int_{x_1}^{x} f'(t)dt \right|\leq | \int_{x_1}^{x}  \left| t-c \right| dt |\leq \frac{1}{2}(c-x_1)^2+\frac{1}{2}(x-c)^2\leq 1 $.
