I'd like to prove that given $f(t,x)$ Lipschitz on $x$ variable for every $[t_1,t_2]x\mathbb{R}^n$. The solution to the IVP is global (defined on $\mathbb{R}$). (with inicial value $x(t_0)= x_0$)
My approach:
Let $[t_1,t_2]$ an interval, I'll prove that the solution is defined on $[t_0,t_2]$. Let $J$ be the maximal interval on the solution and $x_m$ the solution which is defined on this interval, and let $S$ be its supremum. If $t_2 \in J$ we're done. So, let's suppose that it's not.
$s \leq t_2$ and let $t_3= s-\frac{1}{4L}$ beeing $L$ lipschitz constant. Let $y$ be the solution to the IVP with inicial value $y(t_3)=x_m(t_3)$ By existence theorem $y$ is defined on an interval $ [t_3-\frac{1}{2L},t_3+\frac{1}{2L}]\subset I$.
But as $J$ was maximal $t_3+\frac{1}{2L} \in J $ but $t_3+\frac{1}{2L}= s + \frac{1}{4L} < s$. ABS!
Then $t_2 \in J$ and the same argument gives that the solution is defined in $[t_1,t_2]$.
As this works for every interval, then the solution is global for every compact and that implies that the solution is global.