Let $u^\mu$ be a relativistic four-velocity satisfying $u^\mu u_\mu=1$. With Levi-Civita tensor $\epsilon_{\mu\nu\rho\sigma}$, evaluate $\epsilon_{\mu\nu\rho\sigma}u^\mu u^\nu u^\rho u^\sigma$.
The product of four copies of the same vector is symmetric under interchange of any two indices. The Levi-Civita tensor is antisymmetric under such an interchange. Relabelling μ and ν therefore changes the contraction’s sign while leaving its value unchanged. It equals its own negative and must be zero. The normalization of u is not needed for this conclusion.
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