ETH states that for a system, all of its eigenstates thermalize. To be more specific, consider an energy eigenstate of the full system $H|n\rangle=E_n|n\rangle$. If the full system is in this eigenstate, $\rho=\rho^{(n)}=|n\rangle\langle n |$. Denote the system we are interested in as $A$ and the rest of the system as $B$, which is the environment. The reduced density matrix of $A$ is $\rho_A^{(n)}=\mathrm{Tr}_B\left(|n\rangle\langle n |\right)$. ETH states that $\rho_A^{(n)}$ looks thermal: $\rho_A^{(n)}=\rho_A^{\mathrm{eq}}(T_n)$, where $\rho^{\mathrm{eq}}(T) = Z^{-1}\exp(-H/k_BT)$ is the thermal equilibrium density matrix. $T_n$ is deternmined by the energy (density) of the eigenstate $E_n$.
Now, consider the ground state of the full system $|0\rangle$.
Since it is a ground state, the full system must be the lowest temperature possible, which is the zero temperature. Hence $\rho_A^{(0)}=\rho_A^{\mathrm{eq}}(T_0)$ is a thermal equilibrium at $T=0$, which is a pure state with only the ground state.
However, suppose $|0\rangle$ is not a product state (which is often the case, for example, the superfluid ground state of Bose-Hubbard model) so that the reduced density matrix of $A$ is a mixed state.
Since $\rho_A^{(0)}$ cannot be both pure and mixed, a contradiction occurs. What is wrong with the previous argument?