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In the thread above, the proof is based on induction on the dimension of the vector space. However, I think that the induction is valid if you have shown that this bilinear form is also nondegenerate on the orthogonal complement of $\text{span}(x)$, which is the step I cannot think of the proof.
Assume that $x$ has $B(x,x)\neq 0$, then $V=Span(x)\oplus U$ where $U$ is the orthogonal complement of $Span(x)$. Suppose the form was degenerate on $U$ and let $v$ be a degenerate vector for the form on $U$. I claim that $v$ is degenerate on all of $V$. To see this note that by the direct sum decomposition above we can write any $w\in V$ as $tx+u$ where $u\in U$ and $t\in \mathbb{R}$. Then we have
$$B(v,w)=B(v,tx+u)=tB(v,x)+B(v,u)=0$$
where the first term is zero becuase $v$ is in the orthogonal complement of $Span(x)$ and the second term is zero since $v$ is degenerate on $U$. Therefore if $B$ is non-degenerate on $V$ it is also non-degenerate on $U$.
