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

$f(x) = \sqrt{|x|}$ is a famous example of a function which is not Lipschitz continuous but is uniformly continuous. This link shows detailed explanation of it.

Here provides the figure of this function:

However, I am still confused about how to show $\sqrt{|x|}$ is not Lipschitz continuous?

1. Consider $[-a,a]$, which is compact. $\frac{f(y)-f(x)}{y-x}$ should be bounded by $L$. This method is also used to prove $f(x)$ is a uniformly continuous.
2. On $[a,\infty)$ and $(-\infty,-a]$, $f(x)$ has a bounded derivative.

So, based on 1 and 2, $f(x) = \sqrt{|x|}$ is Lipschitz continuous.

I have no idea how to prove it is not a Lipschitz continuous; clearly speaking I do not know how to distinguish the proof of Lipschitz continuity from uniformly continuity.

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 $+\infty$.

The function you seem to consider in 1. is not even defined everywhere. What should it be for $x=y$? Thus this argument does not work.

The "problem" occurs around $0$. Assume the function is Lipschitz with constant $C$. Then $|f(x)-f(0)| \le C |x-0|$ for all $x$. So $f(x)/|x| = \sqrt{|x|}/|x| \le C$ for all $x$ which is not true.