Theorem about non extendable solution of an ODE Cauchy problem

I' ve found in a book without a proof the following theorem: Let $$A=I\times \mathbb{R}$$, where I=]a,b[ is an interval with $$-\infty\leq a, be a set in which the function f(x,y): $$(x,y) \subset A \rightarrow \mathbb{R}$$ is defined, continous and locally lipschitzian in y uniformly in x . Let $$\begin{cases} y'(x)=f(x,y) \\ y(x_0)=y_0 \end{cases}$$ be a Cauchy problem with $$x_0\in I$$. If $$\tilde{y}$$ is a not exctendable unique solution of the problem defined in a maximal interval $$]a_{m},b_m[$$ then we have that:
(i)$$b=b_m$$ or $$b_m and $$\lim_{x \to b_m^{-}} |\tilde{y}(x)| = +\infty$$
(ii)$$a=a_m$$ or $$a_m>a$$ and $$\lim_{x \to a_m^{+}} |\tilde{y}(x)| = +\infty$$
Now I ask where can I find the proof of this theorem and the possible usability in finding the maximal interval of existency of the solution of a Cauchy problem.

A more general statement is that for any ODE IVP $$y'(x)=f(x,y(x)),~~y(x_0)=y_0$$ where $$f$$ has a (open) domain $$D\subset \Bbb R\times \Bbb R^n,~~ (x_0,y_0)\in D,$$ and $$f$$ continuous and locally Lipschitz in $$y$$ on $$D$$, ... , then the (unique) maximal solution leaves any compact set inside of $$D$$.

The proof is essentially that any solution that ends in a compact set has a limit point, and can thus be continued by the local solution of the IVP starting at the limit point.

In your situation, take the compact sets $$[a+\delta,b-\delta]\times[-N,N]\subset D$$.