I'm trying to prove that if $x_n \rightarrow x$ then $\sqrt{x_n} \rightarrow \sqrt x$ with $x_n \geq 0$ and $x \neq 0$.
Applying difference of squares,
$$\left|\, \sqrt{x_n} - \sqrt{x}\,\right| \ = \ \frac{\mid x_n - x \mid}{\sqrt{x_n} + \sqrt x}$$
I've seen that, to bound the denominator (worst case), one can use $\sqrt{x_n} + \sqrt x \geq \sqrt x$. But I was wondering if I can bound $\sqrt x_n$ with the following procedure:
Since $x_n\rightarrow x$ then there exists $N_1$ such that $|x_n - x | < \frac{|x|}{2}$ for all $n\geq N_1$, implying that $\mid x_n \mid > \frac{\mid x \mid}{2}$, and thus $\sqrt{\mid x_n \mid} > \sqrt{\frac{\mid x \mid}{2}}$. Using this fact, I can obtain $$\frac{\mid x_n - x \mid}{\sqrt x_n + \sqrt x} \ < \ \frac{\mid x_n - x \mid}{\sqrt{\frac{x}{2}} + \sqrt x}$$ and finally choose an appropiate $N_2$ for the numerator to make the whole expression less than $\epsilon$.
Are these bounds and the process correct?
Since $\vert x_n - x\vert = \vert (\sqrt{x_n} - \sqrt{x})(\sqrt{x_n} + \sqrt{x})\vert $, it holds that $$\vert \sqrt{x_n} - \sqrt{x}\vert = \frac{\vert x_n - x\vert}{\sqrt{x_n} + \sqrt{x}}$$. Note that the denominator will always be positive so no need to take the absolute value. Furthermore, it is for the same reason true that $\sqrt{x_n} + \sqrt{x}\leq \sqrt{x}$, so
$$\vert \sqrt{x_n} - \sqrt{x}\vert \leq \frac{\vert x_n - x\vert}{\sqrt{x}}.$$ Now, as you have correctly stated, since $x_n$ converges to $x$, there exists for each $\varepsilon>0$ a $N\in\mathbb N$ such that $n\geq N$ implies that $\vert x_n - x\vert<\varepsilon$.
Let $\epsilon>0$, and put $\varepsilon = \sqrt{x}\epsilon$. Note that $x$ does not depend on $n$. Then, by the above, $$\vert\sqrt{x_n} - \sqrt{x}\vert \leq \frac{\vert x_n - x\vert}{\sqrt{x}} < \frac{\varepsilon}{\sqrt{x}} = \frac{\sqrt{x}\epsilon}{\sqrt{x}} = \epsilon.$$ That is, for each $\epsilon>0$, there exists a $N\in\mathbb N$ such that $n\geq N$ implies that $\vert\sqrt{x_n} - \sqrt{x}\vert<\epsilon$, or $\sqrt{x_n}\rightarrow\sqrt{x}$.