Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$

If $x_n>0$ then $x_{n+1}$ cannot be negative. Since $x_1=2>0$, the claim follows.

As, @CiaPan said in comments, you need to prove that the limit to this sequence is exactly $\sqrt 2$.

So, here is a different approach to reach the goal.

Firstly, by induction, prove that $$\color{blue}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^n}}.$$

So, firstly, for $n=1$, we have, $$x_2=\frac 12\left(x_1+\frac 2{x_1}\right)\\\implies\frac {x_2}{\sqrt 2}=\frac {x_1^2+2}{2\sqrt2 x_1}.$$

Now, use inverse of componendo-dividendo, $$\frac {x_2-\sqrt 2}{x_2+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^2.$$

So, for $n=1$, our claim holds.

Now, suppose, it is true for $n=k$, i.e. $$\frac {x_{k+1}-\sqrt 2}{x_{k+1}+\sqrt 2}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^{k+1}}.\tag 1$$

Let's see for $n=k+1$. So, we have $$x_{k+2}=\frac 12\left(x_{k+1}+\frac 2{x_{k+1}}\right),$$ and similar approach to the case $n=1$ takes us to $$\frac {x_{k+2}-\sqrt 2}{x_{k+2}+\sqrt 2}=\left(\frac{x_{k+1}-\sqrt 2}{x_{k+1}+\sqrt 2}\right)^2=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^{k+2}}[from\;\;(1)].$$

So, for all $n$, we have $$\color{blue}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}}=\left(\frac{x_1-\sqrt 2}{x_1+\sqrt 2}\right)^{2^n}=\color{blue}{\left(\frac{2-\sqrt 2}{2+\sqrt 2}\right)^{2^n}}[\text{since }\, x_1=1].$$

So, as $\color{red}{n\to \infty}$, the right side, i.e. $\color{red}{\left(\frac{2-\sqrt 2}{2+\sqrt 2}\right)^{2^n}\to 0}$, so, $$\color{red}{\frac {x_n-\sqrt 2}{x_n+\sqrt 2}\to 0},$$ so $$\color{red}{x_n-\sqrt 2\to 0\\\implies x_n\to \sqrt 2}. $$

So, $${\color{fuchsia}{\lim x_n= \sqrt 2}}$$