The local Lipschitz criterion is the following:
Let $c>0$ and $f \in C([a,b] \times [y_0-c, y_0+c])$.
If $f$ satisfies in $[a,b] \times [y_0-c,y_0+c]$ the Lipschitz criterion as for $y$, uniformly as for $t$,
$$\exists L \geq 0: \forall t \in [a,b] \ \forall y_1, y_2 \in [y_0-c,y_0+c]:$$
$$|f(t,y_1)-f(t,y_2)| \leq L|y_1-y_2|$$
then the ODE $(1) \left\{\begin{matrix} y'(t)=f(t,y(t))\\ y(a)=y_0 \end{matrix}\right.$ is solved uniquely, at least at the interval $[a,b']$
where $A=\max_{a \leq t \leq b , y_0-c \leq y \leq y_0+c} |f(t,y)| \ $ and
$b'=\min \{ b, a+ \frac{c}{A}\}$.
Remark: The continuity of $f, f \in C([a,b] \times \mathbb{R})$ suffices to ensure the existence of a solution of the ODE $(1)$ at an interval $[a,c], c>a$. But, it doesn't ensure us the uniqueness.
For example, $f(y)=\sqrt{|y|}$ doesn't satisfy the local criterion of Lipschitz as for $y$ at none interval that contains $0$.
I tried to show the latter as follows:
$$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|\sqrt{|y_1|}-\sqrt{|y_2|}||\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{|\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{\sqrt{|y_1|}+\sqrt{|y_2|}}$$
Is it right so far? And how do we justify that $f$ doesn't satisfy the local condition of Lipschitz as for $y$ at none of the intervals that contains $0$.
Does it hold because of the fact that it can be that $y_1=y_2=0$?
Also how can we find the intervals at which the local Lipschitz condition is satified?
First question
Since $y_1$ and $y_2$ are as close to $0$ as you want, it is impossible to have $1/(\sqrt{|y_1|}+\sqrt{|y_2|})$ bounded by a constant.
Second question
Most of the times, the easiest way to show that $f(x,y)$ satisfies a Lipschitz condition is to show that $\partial f/\partial y$ is bounded. In this case (for $y>0$) $f'(y)=1/(2\,\sqrt{y})$, which is not bounded as $y\to0$.