Given that $v,w\in V$ and $B$ is nondegenerate bilinear form on V
I showed that ${v}^\perp=\operatorname{span}(v)^\perp$ and ${w}^\perp=\operatorname{span}(w)^\perp$. How to show that $v^\perp=w^\perp\implies \operatorname{span}(v)=\operatorname{span}(w)$.
Please help.
Since $B$ doesn't appear anywhere after being introduced, I'll assume that you intended to imply that orthogonality refers to $B$-orthogonality.
$a=\langle v,v\rangle w-\langle v,w\rangle v$ is orthogonal to $v$. Since $v^\perp=w^\perp$, it's also orthogonal to $w$. Being a linear combination of $v$ and $w$, it is therefore orthogonal to itself. Since $B$ is non-degenerate, $a=0$. Thus $v$ and $w$ are multiples of each other, so either they have the same span or one of them is zero and the other isn't. That last alternative is impossible, since their orthogonal complements coincide.