= Solution
A <vector field> on the <sphere> means a tangent <vector field>. For $S^{2m}\subset\mathbb R^{2m+1}$ with $m\ge1$, suppose $V(x)$ were continuous and nowhere zero. Normalize it to $w(x)=V(x)/\|V(x)\|$. Tangency means $\langle x,w(x)\rangle=0$, and both vectors have norm one. Therefore
$$
H(x,t)=\cos(\pi t)x+\sin(\pi t)w(x),\qquad 0\le t\le1,
$$
has 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 $\operatorname{sgn}\det(-I)=(-1)^{n+1}$. For $n=2m$ its degree is $-1$. <Homotopy invariance of mapping degree> would force $1=-1$, a contradiction. Thus \b[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.
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