Let $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + \sqrt{a_n}}$. Now I want to show by induction that $\sqrt{2} \leq a_n \leq 2$ for all $n$.
The base case is $n=1$ and it is clear $\sqrt{2} \leq a_1 \leq 2$. Then I assume that $\sqrt{2} \leq a_n \leq 2$ holds and I want to show $\sqrt{2} \leq a_{n+1} \leq 2$.
Then I note that $\sqrt{2} \leq a_{n+1} \leq 2 \implies \sqrt{2} \leq \sqrt{2 + \sqrt{a_n}} \leq 2$. By squaring both sides I get $2 \leq 2 + \sqrt{a_n} \leq 4$. Then by subtracting $2$, I get $0 \leq \sqrt{a_n} \leq 2$. This means that $0 \leq a_n \leq 4$. This is ok because I assumed $0 \leq \sqrt{2} \leq a_n \leq 2 \leq 4$.
Edit How about this: Since I know $\sqrt{2} \leq a_n \leq 2$. Then it is clear that $0 \leq a_n \leq 4$. By taking square roots I get $ 0 \leq \sqrt{a_n} \leq 2$. Now if I add 2, $2 \leq \sqrt{a_n} + 2 \leq 4$. Taking another square root I get $\sqrt{2} \leq \sqrt{\sqrt{a_n} + 2} \leq 2$. So $\sqrt{2} \leq a_{n+1} \leq 2$.