Smooth vector fields form a Lie algebra under the Lie bracket

By [L1], the commutator $[X,Y]$ is locally a smooth vector field. Its local expressions agree on chart overlaps because they act as the same operator [L2] on smooth test functions, hence glue uniquely to a global smooth vector field. Thus the bracket closes on $\mathfrak{X}(M)$; no simultaneous choice of charts is needed.

Bilinearity and antisymmetry follow from the corresponding identities for commutators of $\mathbb{R}$-linear endomorphisms of $C^{\mathrm{\infty }}(M)$: $[aX+bY,Z]=a[X,Z]+b[Y,Z],[X,Y]=-[Y,X]\mathrm{.}$

The operator commutator satisfies the Jacobi identity $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ on $C^{\mathrm{\infty }}(M)$, again by direct expansion in the endomorphism algebra.

Steps 1.1-1.3 are exactly the Lie-algebra axioms, so smooth vector fields form a Lie algebra under the Lie bracket.