regular point for smooth vector field is open Let $M$ be a smooth manifold,regular point set for a smooth vector field is open.
My attempt:
Since vector field $X:M\to TM$ is a smooth map,take $X^{-1}(TM\setminus (\bigcup_{p\in M} 0_{T_pM}))$ is open then we are done.
The question then is why $\bigcup_{p\in M} 0_{T_pM}$ is closed in $TM$.So I try to explain it by reviewing the topology on tangent bundle.Which is given by the local chart $TU \to \varphi(U)\times \Bbb{R}^n$ such that open in $TU$ if and only if open in RHS.If one set on $TM$ can be written as union of those open set on $TU$,then it's open in $TM$.So we are done.By checking it on each $TU$
The question is:Is there some better proof for regular point is open set.Or is there some better proof for $\bigcup_{p\in M} 0_{T_pM}$ is closed in $TM$?
 A: I believe the subset of zero vectors that you mean is
$$
Z = \bigsqcup_{p\in M} \{{0_{T_pM}}\} \subset TM = \bigsqcup_{p \in M} T_pM.
$$
The topology of $TM$ defined by basis $\{\varphi_i^{-1}(U_i \times \mathbb{R}^n)\}$ (with $\varphi_i$ is the local trivialization maps) or equally the set $\{\pi_i^{-1}(U_i)\}$.
To show $Z$ is closed, let's try to show instead that $Z^c = TM \smallsetminus Z$ is open. Let $v \in Z^c$ and $\pi_i^{-1}(U_i)$ is an open set contain $v$. Since $\pi_i^{-1}(U_i)$ diffeomorphic with $U_i \times \mathbb{R}^n$, the subset $\varphi_i^{-1}\big(U_i \times (\mathbb{R}^n \smallsetminus\{0\})\big)$ is a open neighbourhood of $v$ contain in $Z^c$. So $Z^c$ open.
So as you said the regular set $X^{-1}\Big(TM \smallsetminus \bigsqcup_{p\in M} \{{0_{T_pM}}\} \Big)$ is open. Another result (its in Lee's Smooth Manifold) that allow us to intuitively see that regular set is open, is this one :

Theorem 9.22 - Canonical form Near a Regular Point : Basically say that if $p\in M$ is a regular point of a smooth vector field $V$ on a smooth manifold $M$, then there exist smooth coordinates $(s^1, \cdots, s^n)$ around $p$ where $V$ represented as
$$
\frac{\partial}{\partial s^1}.
$$

This means that the behaviour of smooth vector field near any regular point is very casual. We can immidiately see that every regular point $p$ surround by other regular points that form an open subsets : on some chart containing $p$, the vector field $V$ is just $\partial/\partial s^1$, which is definitely non-zero.
