Proof for continuity of $\sqrt{x}$ - where's my mistake?

Question

I'm trying to prove that $f(x) = \sqrt{x}$ is continuous on $[0, \infty)$. Here's what I've got so far:

Consider $x_0 \in [0,\infty)$ $$|f(x) - f(x_0)| = |\sqrt{x} - \sqrt{x_0}|$$

Let's multiply that quantity by the positive number $|\sqrt{x}+\sqrt{x_0}|$. Then, $$ \begin{align} |\sqrt{x} - \sqrt{x_0}| &< |\sqrt{x} - \sqrt{x_0}||\sqrt{x} + \sqrt{x_0}|\\ |\sqrt{x} - \sqrt{x_0}| &< |x-x_0| < \delta \end{align} $$

Let's pick $\delta=\epsilon$. Now we have $$ \begin{align} |\sqrt{x} - \sqrt{x_0}| &<\delta \\ |\sqrt{x} - \sqrt{x_0}| &<\epsilon \\ |f(x) - f(x_0)| & < \epsilon \end{align} $$ When $|x-x_0|<\delta$. Seems proven if you ask me, but from what I've read, the modulus of continuity of $\sqrt{x}$ is $\delta(\epsilon)=\epsilon^2$. This forces me to conclude that I've made a mistake somewhere, but I can't find it. (I bet it's some dumb detail)

Thank you in advance!

Answer 1

Hint:

$$|\sqrt x-\sqrt{x_0}|=\frac{|x-x_0|}{\sqrt x+\sqrt{x_0}}<\frac\delta{\sqrt x+\sqrt{x_0}}$$

and you know $\;\sqrt x\ge 0\;$ for all $\;x\in[0,\infty)\;$ , so $\;\sqrt x+\sqrt{x_0}\ge\sqrt{x_0}\;$ ...