Let $\psi: \mathbb{C} \to \mathbb{R}$ be a bijection (such a function exists since $|\mathbb{C}|=|\mathbb{R}^2|=|\mathbb{R}|$).
Define a binary relation $R$ on $\mathbb{C}$ such that $z_1Rz_2$ iff $\psi(z_1)<\psi(z_2)$.
Then $R$ is transitive: if $z_1Rz_2$ and $z_2Rz_3$ then $\psi(z_1)<\psi(z_2)$ and $\psi(z_2)<\psi(z_3)$, so $\psi(z_1)<\psi(z_3)$, hence $z_1Rz_3$.
$R$ is strict: if $z_1Rz_1$ then $\psi(z_1)<\psi(z_1)$, impossible.
$R$ is trichotomous: either $\psi(z_1)<\psi(z_2)$,$\psi(z_2)<\psi(z_1)$, or $\psi(z_1)=\psi(z_2)$, therefore either $z_1Rz_2$, $z_2Rz_1$ or $z_1=z_2$.
Thus $R$ is a strict total order on $\mathbb{C}$.
However I was told there does not exist a total order on $\mathbb{C}$.
So what is the error here?
A total ordering exists on $\mathbb C$, as you have proved. However, there exists no total ordering on $\mathbb C$ which "makes sense together with its other structure". By that, we usually mean that $\mathbb C$ does not admit an ordering which turns it into an ordered field. That is, it needs to be translation invariant, and any two positive numbers (that is, $\geq 0$) multiplied by one another are positive again.
From the translation invariance of an ordered field, you get $a \leq 0 \iff 0 \leq -a$, so it follows that for any $a \neq 0$ you have that precisely one of $a,-a$ is positive. Now you can prove: precisely one of $-1,1$ is positive; by squaring this number, you see that $1$ must be the positive one; precisely one of $i,-i$ must be positive; but then by squaring this number $-1$ must be positive. Contradiction.