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

 Let $x\in (0,2)$. Since, $f\left(x\right)$ satisfies the requirements of Lagrange's mean value theorem in $[0,2]$. 

So, it also satisfies in $[0,x]$. Consequently, there exists $c\in (0,x)$ such that $\left|\frac{f\left(x\right)-f\left(0\right)}{x-0}\right|=f\text{'}\left(c\right)$

$\Rightarrow \left|\frac{f\left(x\right)}{x}\right|=\left|f^{\text{'}}\left(c\right)\right|\le \frac{1}{2}$

$\Rightarrow \left|f\right(x\left)\right|\le \frac{x}{2}$

Since $0<x<2$

$\Rightarrow \left|f\right(x\left)\right|\le 1$

Hence, option-$3$ is the correct answer.