Given,

$f\left(x\right)=\max \left\{1+x+\left[x\right], 2+x, x+2\left[x\right]\right\}, 0\le x\le 2$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}\max \left\{x+1, x+2, x\right\}& 0\le x<1\\ \max \left\{x+2, x+2, x+2\right\}& 1\le x<2\\ \max \left\{5, 4, 6\right\}& x=2\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{cc}x+2,& 0\le x<1\\ x+2,& 1\le x<2\\ 6,& x=2\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{cc}x+2& 0\le x<2\\ 6& x=2\end{array}\right.$

So, $f$ is not continuous at $x=2$ as $f\left(2^{-}\right)=4 \& f\left(2\right)=6$

And $f$ is differentiable in $\left(0, 2\right)$ as $x+2$ is a linear function,

$\therefore m=1, n=0$

$\Rightarrow {\left(m+n\right)}^2+2=1+2=3$

Hence this is the required option.