A recursive sequence converging to 2

Bounded. We show $0\le u_n<2$ by induction. It holds for $n=0$. If $0\le u_n<2$, then $2\le2+u_n<4$, so $\sqrt2\le u_{n+1}<2$.

Increasing. Since all terms are non-negative,
$$u_{n+1}^2-u_n^2=2+u_n-u_n^2=(2-u_n)(1+u_n)>0,$$
so $u_{n+1}>u_n$.

Limit. An increasing sequence bounded above converges, say to $L$ with $0\le L\le2$. Passing to the limit in $u_{n+1}=\sqrt{2+u_n}$ gives $L=\sqrt{2+L}$, so $L^2-L-2=0$, that is $(L-2)(L+1)=0$. Since $L\ge0$, $L=2$.