Let F defined on [0,1] be twice differentiable

As $f(0)=f(1)$ and $f$ is differentiable, hence by Rolles theorem:

$f^{\prime}(c)=0$ for some $c \in[0,1]$

let us now apply LMVT (as function is twice differentiable) for point $c$ and $x \in[0,1]$,

hence,

$\frac{\left|f^{\prime}(x)-f(c)\right|}{x-c}=f^{\prime \prime}(d)$

$\Rightarrow \frac{|f(x)-0|}{x-c}=f^{\prime \prime}(d)$

$\Rightarrow \frac{|f(x)|}{x-c}=f^{\prime \prime}(d)$

A given that $\left|f^{\prime \prime}(d)\right|<=1$ for $x \in[0,1]$

$\Rightarrow \frac{|f(x)|}{x-c} \leq 1$

$\Rightarrow|f(x)| \leq x-c$

Now both $x$ and $c$ lie in $[0,1]$, hence $x-c \in[0,1]$