Proof of Non-Ordering of Complex Field

To show that $-1$ is not in $P$, note that if $-1\in P$ then $(-1)(-1)\in P$, which contradicts the fact that if $x \ne 0$ exactly one of $x$ and $-x$ is in $P$.

Next we show that $i\notin P$. Suppose to the contrary that $i\in P$. Then $i^2\in P$, which contradicts the fact that $-1\notin P$.

The same argument shows that $-i\notin P$. This contradicts the fact that if $x\ne 0$, then exactly one of $x$ and $-x$ is in $P$.

I think the definition of $P$ you have is slightly off: if $x=0$ then all three conditions are satisfied. A possible fix is "Either $x\in P$ or $-x\in P$, with both holding iff $x=0$." On the other hand, I'm not convinced that it's important that $0\in P$; you should check that before making things complicated.

For an arbitrary field, $-1\notin P$ because then $(-1)(-1) = 1\in P$, which is impossible.

From there, you assume that $P$ exists and begin a proof by contradiction. Using your work in the OP you can therefore show that $i\notin P$. However, since $i\neq 0$, we also need to show that $-i\notin P$ before we continue. The proof is essentially identical to the one you gave in the OP.

This will contradict the fact that either $i$ or $-i$ is in $P$. Therefore, there cannot be such a set $P\subset\mathbb{C}$.