Assume that $z_1, z_2, z_3, z_4$ lie on a line or circle C.
By what I just mention above, there exist a mobius tran $T$ that takes the real numbers $x_2, x_3, x_4$ to $z_2, z_3,z_4$ respectively. Since there is only one such line or circle and since we know that Mobius take $\mathbb{R}_{\infty}$ to a line or circle, we see that there exist a real number $x_1$ such that $T(x_1) = z_1$.
Since the cross ratio is preserved under a Mobius trans, then $ (x_1,x_2,x_3,x_4)=(T(x_1),T(x_2),T(x_3),T(x_4))=(z_1,z_2,z_3,z_4)$
But since $x_i$ is all real numbers, its cross ratio is also real, which show that the cross ratio of $z_i$ is also real.
Assume that $z_1$, $z_2$, $z_3$ and $z_4$ lie on a line or circle $C$. Let $\phi$ be the Möbius transformation taking $1, 0, \infty$ to $z_2$, $z_3$, $z_4$, respectively. The image of $\mathbb{R} \cup \{\infty\}$ under $\phi$ is a line or circle passing through $z_2$, $z_3$ and $z_4$. Since there is only one such line or circle, it must be $C$. Therefore $\phi^{-1}(z_1) \in \mathbb{R} \cup \{\infty\}$. But $(z_1, z_2, z_3, z_4) = \phi^{-1}(z_1)$.