Prove that any solution of an equation $\dot x=\mathbf v(t,x)$ defined by a direction field in $\Bbb R\times\Bbb R^n$ can be extended indefinitely if $\mathbf v$ grows no faster than the first power of $x$ at infinity, i.e., if $|\mathbf v(t,x)|\le k|x|$ for all $t$ and all $|x|\ge r$, where $r$ and $k$ are constants.
Hint. Comparing with a motion in the field $\dot x=kx$, construct comppact sets whose boundaries require arbitrarily long times to reach.
I'm puzzled about how to construct it.
Obviously, if a solution does not leave the ball of radius $r$ you are done. For any segment of a solution outside $r$ we get the estimate of the radius evolution of $$ u=\|x\|\implies \dot u = \frac{x^T\dot x}{\|x\|}\le\|\dot x\|=\|v\|\le k\|x\|=ku $$ So to reach, if at all, the radius $2r$ starting from the point crossing the sphere of radius $r$ would need time $$ t(2r)-t(r)=\int_{t(r)}^{t(2r)}\,dt=\int_r^{2r}\frac{du}{\dot u}\ge\int_r^{2r}\frac{du}{ku}=\frac{\ln2}{k} $$ This holds also for all later radius doublings, thus in finite time one can only reach a finite radius.