Going to detail what I said in my comments:
You seem to be confused about which implication is trivial or not.
It is true that for a fixed $v$, the property "$(A):\,\,\, \forall w \in V,\, \langle v,w\rangle = 0$" implies that $\langle v,v\rangle = 0$ by taking $w = v$.
However what that means is that, if $\langle\cdot,\cdot\rangle$ is positive-definite, then, given any $v$ satisfying $(A)$, we get that $v = 0$, which provides "positive-definite implies nondegenerate" and not the other way round!
As for the other implication, let me first detail the proof for Cauchy-Schwarz when the form isn't positive-definite:
Lemma ("generalised" Cauchy-Schwarz): Let $\langle\cdot,\cdot\rangle : V \to \mathbb{R}$ be a positive symmetric bilinear form on $V$ a $\mathbb{R}$-vector space.
Then, we have:
$$\forall (v,w) \in V^2,\quad \langle v,w\rangle^2 \leq \langle v,v\rangle\cdot\langle w,w\rangle$$
Proof: Let $(v,w) \in V^2$. Denote by $P$ the following polynomial function:
$$P : t \in \mathbb{R} \mapsto \langle v + tw,v+tw \rangle \in \mathbb{R}$$
After expanding and using the symmetry, we have:
$$P(t) = \langle v,v\rangle + 2t\langle v,w\rangle + t^2 \langle w,w\rangle$$
Yet by positivity of $\langle \cdot,\cdot \rangle$ $P$ is positive.
There are two cases now:
- If $\langle w,w\rangle \neq 0$: then the proof is just like regular Cauchy-Schwarz. $P$ is a positive quadratic polynomial, thus its discriminant is nonpositive, which provides what we want.
- If $\langle w,w\rangle = 0$: $P$ is then a positive affine function, thus its leading coefficient $\langle v,w\rangle$ must be $0$, which does give what we desire since $0^2 \leq \langle v,v\rangle \cdot 0$.
QED (for the lemma)
Now, let's suppose our $\langle \cdot,\cdot \rangle$ is non-degenerate. Let $v \in V$ be such that $\langle v,v\rangle = 0$.
Then, for all $w \in V$, by Cauchy-Schwarz:
$$0 \leq \langle v,w \rangle^2 \leq \langle v,v\rangle \cdot \langle w,w\rangle = 0$$
Thus: $\forall w \in V,\, \langle v,w\rangle = 0$, thus, by non-degeneracy: $v = 0$, and $\langle \cdot,\cdot\rangle$ is positive-definite, hence the equivalence by double implication.