If $x_{n+1}-\alpha=\frac{(x_{n}-\alpha)^2}{2x_{n}}$
Show that for n >1, $$|{x_{n+1}-\alpha}|\le\frac{1}{2}|x_{n}-\alpha|$$
$|{x_{n+1}-\alpha}|=|\frac{(x_{n}-\alpha)^2}{2x_{n}}|=\frac{1}{2}|\frac{(x_{n}-\alpha)^2}{x_{n}}|=\frac{1}{2}|\frac{x_{n}^2-2x_{n}\alpha+\alpha^2}{x_{n}}|$ $=\frac{1}{2}|x_{n}-2\alpha+\frac{\alpha^2}{x_{n}}|$
I'm stuck at this point - don't know how to manipulate it further to reach the inequality
Actually, such relation should hold only if you add the assumption $\alpha \geq 0$. I suggest you to argue as follows: you want to prove that $$x_{n+1} - \alpha=\frac{(x_n - \alpha)^2}{2x_n}\leq \frac{|x_n-\alpha|}{2},$$ multiplying by $2x_n$, and dividing by $|x_n-\alpha|$, you find out that this is EQUIVALENT to prove that $$x_n - \alpha\leq x_n,$$ that is, $\alpha \geq 0$, bu that was our assumption, and so we are done.
You see that, since we got equivalent conditions to our claim, you actually need $\alpha \geq 0$. Does it seem clear?