 Solution

A vector field on the sphere means a tangent vector field. For $S^{2m}\subset {\mathbb{R}}^{2m+1}$ with $m\ge 1$, suppose $V(x)$ were continuous and nowhere zero. Normalize it to $w(x)=V(x)/\Vert V(x)\Vert$. Tangency means $⟨x,w(x)⟩=0$, and both vectors have norm one. Thereforehas squared norm ${\cos }^2(\pi t)+{\sin }^2(\pi t)=1$. It defines a homotopy through self-maps of the sphere from the identity to the antipodal map.

The identity has mapping degree $1$. The antipodal map on $S^n$ has degree $(-1{)}^{n+1}$: it is the boundary restriction of $-I$ on the oriented $(n+1)$-ball, and this linear map changes orientation by $\mathrm{sgn}\det (-I)=(-1{)}^{n+1}$. For $n=2m$ its degree is $-1$. Homotopy invariance of mapping degree would force $1=-1$, a contradiction. Thus every continuous tangent field on an even-dimensional sphere vanishes somewhere. For $S^0$ the tangent spaces are zero, so the conclusion holds directly. This proves the Hairy ball theorem using degree, rather than assuming it.