Let $f(x)$ satisfy the requirement of Lagrange's Mean value theorem in $[0,2]$. If $f(0)=0$ and $\left|f'(x)\right| \leq \frac{1}{2}$ for all $x \in[0,2]$, then

$|f(x)| \leq 1$

Let $x \in(0,2)$. Since $f(x)$ satisfies the requirements of Lagrange's mean value theorem in [0,2]. So, it also satisfies in $[0, x]$. Consequently, there exist $c \in(0, x)$ such that

$f'(c) =\frac{f(x)-f(0)}{x-0}$

$\Rightarrow f'(c)=\frac{f(x)}{x}$

$\Rightarrow \left|\frac{f(x)}{x}\right| =\left|f'(c)\right| \leq \frac{1}{2}$          $\left[∵\left|f^{\prime}(x)\right| \leq \frac{1}{2}\right]$

$\Rightarrow |f(x)| \leq \frac{|x|}{2}$

$\Rightarrow |f(x)| \leq \frac{x}{2}$                 $[∵ x>0]$

$\Rightarrow |f(x)| \leq 1$                  $[∵ x \in(0,2) \quad ∴|x|<2]$