This is a very simple fact to prove, but I am worried that my proof is so simple as to have no chance of being right.
For a vector space $V$, $-(-v) = v$ for any $v \in V$.
Proof. Let $v \in V$. By the additive inverse axiom, $$v + (-v) = (-v) + v = 0,$$ so $v$ is the additive inverse of $-v$. Since the additive inverse in a vector space is unique, and we denote the a dditive inverse of $-v$ by $-(-v)$, we have $$v = -(-v).$$
Have I skipped any steps?
Your proof sounds good. As you have mentioned, the additive inverse is unique. To be rigorous, we are going to prove it next.
Indeed, consider a vector $v\in V$ and let us assume it admits two different additive inverses, which we shall denote by $w$ and $z$. Consequently, we have that \begin{align*} w = w + 0 = w + (v + z) = (w + v) + z = 0 + z = z \end{align*} and we are done.
Based on such result, we conclude that \begin{align*} (-v) - (-v) = 0 = (-v) + v \Longrightarrow v = -(-v) \end{align*} as desired. Hopefully this helps.