Let $M$ be a smooth differential manifold, and let $X\in\mathfrak{X}(M)$ be a smooth vector field. The support $supp(X)$ of $X$ is defined as the closure, in $M$, of the set $A(X):=\{m\in M: X(m)\neq 0\}$. I am not sure to be able to correctly ``see'' $supp(X)$ even for simple vector fields. For example, if $M=\mathbb{R}^{2}$ with cartesian coordinates $(x\,;y)$, and $X=y\partial_{x}$, it is evident that $A(X)$ is the complement of the $x$-axis. In this case, since every open neighbourhood of a point on the $x$-axis contains a point in $A$, I think that $supp(X)=M$. However, this line of reasoning leads me to think that if $X$ is globally defined on $M$, then $supp(X)=M$. Is it true?
Furthermore, I am concretely interested in the set of critical points of $X$ defined as $Z(X):=\{m\in M: X(m)=0\}$, and I would like to know if it is closed or not. I am sorry for not having any personal thought on this, but, at the moment, I seem to be unable to come up with anything at all.
Thank You.
Of course not, consider the zero vector field. The picture you should have in your head is pretty clear from the definition - $\mathrm{supp}(X)$ is the smallest closed set that contains all the points where $X$ is non-zero. So the only vector fields whose support is all of $M$ are those that are non-zero on a dense set.
Regarding your second question, remember from topology that since $X$ is continuous, the preimage under $X$ of a closed set is closed.