Given $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $, prove $\sqrt{2} ≤ x_n ≤ 2$ 

Let $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $ for all $n \in \mathbb{N}$.
By using proof by induction:
(i) Prove that $\sqrt{2} ≤ x_n ≤ 2$ for all $n \in \mathbb{N}$.
(ii) Prove that $x_n ≤ x_{n+1}$ for all $n \in \mathbb{N}$.


For (i)
Let's prove the case base case, for $n=1$ we have that $\sqrt{2} \leq x_1\leq 2$ which is clearly true by $x_1=\sqrt{2}$ 
Now we assume true for $n=k$ $\Longrightarrow$ $\sqrt{2}\leq x_k\leq2$ and we are required to prove $\sqrt{2}\leq x_{k+1} \leq 2$.
From our hypothesis : $\sqrt{2}\leq x_k\leq2$ however we know that $x_{k+1} =\sqrt{2x_k}\Longrightarrow x_k = \frac{(x_{k+1})^2}{2}$ so we have  $\sqrt{2}\leq \frac{(x_{k+1})^2}{2}\leq2 \Leftrightarrow \sqrt{2\sqrt{2}}\leq x_{k+1} \leq 2$. 
How do I proceed?

From
$$\sqrt2<x_n<2$$ you can deduce
$$\sqrt{2\sqrt2}<\sqrt{2x_n}<2$$
and obviously
$$\sqrt{2}<\sqrt{2x_n}<2,$$ which is nothing but
$$\sqrt{2}<x_{n+1}<2.$$
