Clearly the question as stated cannot be correct, as I can take the zero bilinear form, and take any nontrivial subspace. But I'm wondering where in the following proof I used nondegeneracy and symmetry, or please point out the mistake.
I'm given the following exercise: $V$ be a finite dimensional vector space equipped with a symmetric bilinear form $B$, and $W$ a subspace such that $B|_W$ is nondegenerate. Then, $V\cong W\oplus W^\perp$. (Recall $W^\perp:=\{v\in V: B(v,w)=0, \forall w\in W\}$)
My proof (attempt?) is to do the following: Take an orthonormal basis $w_1,...,w_k$ for $W$. Consider the projection operator $p:V \to W$ given by $$p(v):=\sum_{i=1}^k B(v,w_i)w_i.$$ p is the identity on $W$ so it is surjective. By the first isomorphism theorem, we have a short exact sequence of vector spaces
$$0\to \ker p \to V\to W\to 0.$$
Since every SES of vector spaces splits, it suffices to show $\ker p= W^\perp$. If $v\in W^\perp$, then in particular for all $i$, $B(v,w_i)=0$. Thus, $$p(v)=\sum B(v,w_i)w_i=0.$$
For the other direction, if $v\in \ker p$, then $$0=\sum B(v,w_i)w_i\implies B(v,w_i)=0$$ since $w_i$ forms a basis for $W$. Then, for an arbitrary $w\in W$, we have $w=\sum_i a^i w_i$, so $$B(v,w)=B(v,\sum a^iw_i)=\sum a^iB(v,w_i)=0.$$
So where did I use nondegeneracy of $B$ when restricted to $W$? And where did I use symmetry of $B$?.