How to show square root of absolute of x, $\sqrt{|x|}$, is not Lipschitz continuous?

The derivative of $\sqrt{|x|}$ is $\frac{\mathbb{sgn}(x)}{2\sqrt{|x|}}$. Let $x_0 = \frac{1}{4L^2}$. The derivative at $x_0$ is $L$, so, the derivative is unbounded.

Consider only the interval $[0,a]$ for some $a>0$ and suppose that $f(x)=\sqrt{|x|}$ is Lipschits: $|f(y)-f(x)|\leq L|y-x|$. Then for $x=0$ and $y<a$ we should have:

$|\sqrt{y}|\leq L|y|\Leftrightarrow \sqrt y\leq Ly\Rightarrow \frac{\sqrt y}{y}\leq L$ for each $y>0$. But this is impossible since for $y\rightarrow 0^+$, the LHS tends to $+\infty$.