Let $x_{0}\in [0,+\infty)$, $\epsilon>0$
Suppose we have: $|x-x_{0}|<\delta$ where $\delta>0$
$|f(x)-f(x_{0})|=|\sqrt{x}-\sqrt x_{0}|\le\sqrt{x}+\sqrt x_{0}$
$|x-x_{0}|<\delta\implies x<\delta+x_{0}\implies \sqrt{x}<\sqrt{\delta+x_{0}}\le\sqrt{\delta}+\sqrt{x_{0}}$
So if we want to make $\sqrt{\delta}+2\sqrt x_0\le\epsilon$, we get $\delta\le (\epsilon-2\sqrt x_{0})^2$, but what if $x_{0}=\epsilon$ ?