I knew this question was asked before, like here Non-vanishing vector fields on non-compact manifolds .
But it seems there is no detailed satisfying answer I could find. So can anyone please give proof that for this result: Any non-compact smooth manifold $M$ has a non-vanishing vector field?
Thank you in advance!
I answered the question you mention and found this one afterwards, I suppose I´ll leave the answer here too.
First of all, you can take a vector field $V_0$ in $M$ with isolated zeros as shown here. The idea of the proof is basically "pushing all the zeros to infinity".
Take a covering of the manifold $M$ by compacts $C_n$, $n\geq1$, with $C_n$ contained in the interior of $C_{n+1}$ $\forall n$. We´ll also need $M\setminus C_n$ not to have any component which is relatively compact in $M$ ($*$). You can achieve this by adjoining to $C_n$ all the relatively compact components of $M\setminus C_n$ (the resulting set is compact, see this).
Now we are going to inductively define a field $V_n$ that has no zeros in $C_n$. To do that, we take the vector field $V_n=(\phi_n)_*V_{n-1}$, where:
By induction hypothesis, $V_{n-1}$ has isolated zeros, and none of them are in $C_{n-1}$.
$\phi_n:M\to M$ is an diffeomorphism fixing $V_{n-1}$ and taking all the (finitely many) zeros of $V_{n-1}$ that were inside $C_n$ outside of it. (To make sure this exists you need the condition ($*$), which implies that every point outside $C_{n-1}$ is connected by a path in $M\setminus C_{n-1}$ to some point outside $C_n$, so you can take it outside $C_n$ by a diffeo fixing $C_{n-1}$).
Now as the sequence $V_n$ of vector fields is eventually constant in any compact subset of $M$, you can consider its limit $V$, which is equal to $V_n$ in $C_n$ so it can´t have any zeros in $C_n$ $\forall n$, meaning it has no zeros.