I do not understand how $\varphi=x_0$ is the complete solution to the IVP $\dot{x}=v(x)$ where $v(x_0)=0$. If for some other x we have $v(x)\neq 0$ then $\varphi=x_0$ obviously does not cut it. So why can this not happen? Does it have something to do with the derivative of the solution being zero at the start?
Any help would as always be greatly appreciated. I do apologise if the answer is very simple, but I am quite frustrated and currently have noone to discuss it with.
I am sorry. The answer for this is given in the proof and we need differentiability for $v$.
Basically it must be the case that $\varphi (t) \equiv x_0$, because if not then around $x_0$ (which we may WLOG set to zero by a translation) we have $|v(x)| < k|x|$ because of the differentiability of the field $v$. The solution to $\dot{x}=kx$ does not reach zero in finite time and so the same must apply to the solution to $\dot{x}=v(x)$, whose rate of change was shown to be slower.
Still, as far as intuition goes it is still quite unclear. But I guess differentiability is the key.