= Solution
The map
$$
\ell_v:V\to\mathbb R,
\qquad
\ell_v(x)=\langle v,x\rangle
$$
is a nonzero <linear functional>: otherwise $v$ would lie in the <radical of a bilinear form>, contradicting nondegeneracy. Its kernel is $v^\perp$, so the <rank-nullity theorem> gives $\dim v^\perp=n-1$. Antisymmetry gives $\langle v,v\rangle=-\langle v,v\rangle=0$, hence $v\in v^\perp$.
Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^\perp$, it is also orthogonal to $v$, and therefore to
$$
W\oplus\mathbb Rv=v^\perp.
$$
For a <nondegenerate bilinear form>, $(v^\perp)^\perp=\mathbb Rv$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap\mathbb Rv=\{0\}$, proving that the restriction to $W$ is nondegenerate.
The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional <symplectic vector space> has even dimension.
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