Let $X=(X_1,\dots, X_n)\in L^\infty$ be a smooth (non Lipschitz) vector field such that
\begin{equation}\tag{1} X_n \ge c|(X_1, \dots, X_{n-1})|\,. \end{equation}
Does the cone condition (1) imply that the flow $\Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.
Is this true?
Take $\bar x$ and given the maximal interval $(0, t_{\bar x})$, I argue by contradiction that $t_{\bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{\bar x}$ and let $x_n$ be the points corresponding to the flow $\Phi(t_n,\bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{\bar x}$ by setting it equal to $\lim_n x_n$.
How do I prove that these $x_n$ stay bounded, given (1)?
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz. You have that $$ |\Phi(t,x)-x| = \left|\int_0^t X(\Phi(s,x))ds\right| \leq \int_0^t |X(\Phi(s,x))|ds \leq \|X\|_\infty t. $$ This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,\infty)$.