Prove that a sequence converges to $\sqrt{2}$

Let's write this $x_{n+1}=\dfrac12\left(x_n+\dfrac2{x_n}\right)$. The sequence $x_n$ has actually a closed form.

First, notice that $x_n>\sqrt{2}$ for all $n$, by induction: if $x_n>\sqrt{2}$, then

$$x_{n+1}-\sqrt2=\dfrac12\left(x_n-2\sqrt{2}+\dfrac2{x_n}\right)=\frac12\left(\sqrt{x_n}-\dfrac{\sqrt2}{\sqrt{x_n}}\right)^2\geq0$$

And it can be equal to $0$ only if $x_n=\sqrt{2}$, which is wrong by hypothesis. Since $x_1>\sqrt{2}$, it's then true for all $n$.

Now, a little bit of hyperbolic trigonometry: $\coth x>1$ for $x>0$, and it's a decreasing function, from $+\infty$ (for $x\to0^+$) to $1$ (for $x\to+\infty$), hence it's a bijection from $]0,+\infty[$ to $]1,+\infty[$.

You can then find, for all $n$, a unique $\mu_n>0$ such that $x_n=\sqrt{2}\coth \mu_n$. Then simplify:

$$x_{n+1}=\dfrac12\left(\sqrt{2}\coth \mu_n+\dfrac2{\sqrt{2}\coth \mu_n}\right)=\frac{\sqrt{2}}2\left(\coth \mu_n+\tanh \mu_n\right)=\sqrt{2}\coth(2\mu_n)$$

Hence $\mu_{n+1}=2\mu_n$, and

$$x_n=\sqrt{2}\coth\left(2^{n-1}\mathrm{arccoth} (x_1)\right)$$

Since $\mathrm{arccoth} (x_1)>0$, you have $2^{n-1}\mathrm{arccoth} (x_1)\to+\infty$ as $n\to\infty$, hence

$$\lim_{n\to\infty} x_n=\sqrt{2}$$