The IVP $y'=x|y|$ is given along with the condition $y(1)=0$. Upon checking the Lipschitz condition one gets,
$|x|*||y_{2}|-|y_{1}||\le|x|*|y_{2}-y_{1}|$
Now, if I go locally around $x=1$, $|x|$ shall keep on increasing ahead of $x=1$.
So, as I see it, I will never get that fixed L in
$||y_{2}|-|y_{1}||\le L|y_{2}-y_{1}|\ \ \ \ \ \ \ \ \forall(x,y)$
to prove the IVP having a unique solution.
The answer in the book says that the IVP does have a unique solution around $x=1$.
Where am I wrong with my argument?
Such an $L$ can be found.
There are 6 cases to check, but due to the symmetries of the problem this is can be reduced to just two case. I will show you those two cases, and leave it up to you to verify the rest.
Case 1: $0 \leq y_1 \leq y_2$ and $|y_1| \leq |y_2|$
We have \begin{align*} ||y_2| - |y_1|| = |y_2 - y_1| = y_2 - y_1, \end{align*} and, $$|y_2 - y_1| = y_2 - y_1.$$ Therefore, the constant $L = 1$ will suffice here.
Case 2: $y_1 \leq 0 \leq y_2$ and $|y_1| \leq |y_2|$ We have $$||y_2| - |y_1|| = |y_2 + y_1| = y_2 + y_1,$$ and, $$|y_2 - y_1| = y_2 - y_1 \geq y_2 + y_1.$$ Again, the constant $L=1$ is sufficient.
From here, all other case are symmetries of these first two cases.
I hope this answers your question. I am not entirely confident I correctly understood your problem.