Proof that $f(x)=\sqrt{x}$ is locally Liptschitz

Let $f:(0, 1] \to \Bbb R$ given by $f(x)=\sqrt{x}$, I want to show that $f$ is locally Liptschitz, this means (for $\lambda>0$, and $x, y \in B(p;r)$)

$$|\sqrt x-\sqrt y| \leq \lambda |x-y|$$If we take $x, y \in B(0, r)$ for any $r>0$, then $|x|<r$ and $|y|<r$. As $x,y \in (0, 1]$ then $|\sqrt{x}|<|x|$ and $|\sqrt{y}|<|y|$ from this maybe I can say that:

$$|\sqrt x-\sqrt y| \leq |\sqrt{x}|+|\sqrt{y}| \leq |x| + |y|=2r$$

But I think this is not my goal as such. Any suggestions to even out the inequality?