Taking root from absolute expression

Question

Why is the following true? (Where all terms are positive)

$$|x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon$$

Answer 1

\begin{align} |\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{|\sqrt{x} + \sqrt{y}|} \le \frac{|x - y|}{|\sqrt{x} - \sqrt{y}|} < \frac{\epsilon^2}{|\sqrt{x} - \sqrt{y}|} \end{align}

Answer 2

An other proof would be:

If $0\leq y\leq x\leq$, $$\left|\sqrt x-\sqrt y\right|^2=\left(\sqrt x-\sqrt y\right)^2=x-2\sqrt x\sqrt y+y\leq x-2y+y=x-y=|x-y|,$$

The proof for $0\leq x\leq y$ is the same. Then for all $x,y\geq 0$, $$\left|\sqrt x-\sqrt y\right|\leq \sqrt{|x-y|}$$

and so, $$|x-y|<\varepsilon^2\implies \left|\sqrt x-\sqrt y\right|<\varepsilon$$

Q.E.D.