Prove that $0v = 0$ for any vector $v \in V$.
I'll assume the existence of that $a(b+c) = ab+bc$, $0\times 0 = 0$, $v-v = 0$ and that every vector has an inverse.
$$0v=0$$
$$0v+(-0v) = -0v$$
$$0(v-v) = 0\times 0 = 0 = -0v $$
Similarly,
$$-0v = 0$$
$$-0v + 0v = 0v$$
$$0(-v+v) = 0\times 0 = 0 = 0v$$
I'm not quite sure if it would be fine to assume the existence of the properties I used in this proof.
It seems like you are assuming that $0v=0$ to prove that $0v=0$, which is a circular reasoning and proves nothing.
The trick is: $$0v=(0+0)v=0v+0v$$ Then add $-(0v)$ and you get $$0=0v$$