My previous attempt to explain the argument wasn't quite successful, so let's try again.
Suppose that there's a ψ-epistemic ontological model for the quantum theory of a single qubit. (The terminology is defined in
HS. See also
ML). Denote the set of ontic states of that model by \Lambda. Then \Lambda\times\Lambda is the set of ontic states in an ontological model for the two-qubit quantum theory.
I'm going to simplify the presentation of the argument by pretending that \Lambda has a finite number of members. (I want to avoid technical details about probability measures). Denote that number by n, and denote the members of \Lambda by \lambda_1,\dots,\lambda_n.
Let \mathcal H be the Hilbert space of the quantum theory of a single qubit. Then \mathcal H\otimes\mathcal H is the Hilbert space of the quantum theory of two qubits. Let \{|0\rangle,|1\rangle\} be an orthonormal basis for \mathcal H. Define
[tex]
\begin{align}
|+\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle+|1\rangle\right)\\
|-\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle-|1\rangle\right).
\end{align}
[/tex] \{|+\rangle,|-\rangle\} is another orthonormal basis for \mathcal H.
For each |\psi\rangle\in\mathcal H and each \lambda\in\Lambda, let Q_\psi(\lambda) denote the probability that the qubit's ontic state is \lambda. The function Q_\psi:\Lambda\to[0,1] is called the epistemic state corresponding to |\psi\rangle. Similarly, for each |\psi\rangle\otimes|\psi'\rangle\in\mathcal H\otimes\mathcal H and each (\lambda,\lambda')\in\Lambda\times\Lambda, let Q_{\psi\psi'}(\lambda,\lambda') denote the probability that the two-qubit system is in ontic state (\lambda,\lambda'). We assume that
[tex]Q_{\psi\psi'}(\lambda,\lambda') =Q_\psi(\lambda)Q_{\psi'}(\lambda')[/tex] for all values of the relevant variables.
Let X be a self-adjoint operator on \mathcal H\otimes\mathcal H with the eigenvectors
[tex]
\begin{align}
|\xi_1\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle\otimes|1\rangle +|1\rangle\otimes|0\rangle\right)\\
|\xi_2\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle\otimes|-\rangle +|1\rangle\otimes|+\rangle\right)\\
|\xi_3\rangle &=\frac{1}{\sqrt{2}} \left(|+\rangle\otimes|1\rangle +|-\rangle\otimes|0\rangle\right)\\
|\xi_4\rangle &=\frac{1}{\sqrt{2}} \left(|+\rangle\otimes|-\rangle +|-\rangle\otimes|+\rangle\right)
\end{align}
[/tex] Note that each of the state vectors
[tex]
\begin{align}
&|0\rangle\otimes|0\rangle\\
&|0\rangle\otimes|+\rangle\\
&|+\rangle\otimes|0\rangle\\
&|+\rangle\otimes|+\rangle
\end{align}
[/tex] is orthogonal to exactly one of the |\xi_k\rangle.
The result of an X measurement that corresponds to eigenvector |\xi_k\rangle will be denoted by k. For all k and all \psi,\psi'\in\mathcal H, let P_{\psi\psi'}(k|X) denote the probability assigned by the ontological model for the two-qubit quantum theory to measurement result k, given that we're measuring X, and that the epistemic state of the two-qubit system is Q_{\psi\psi'}. For each k and each \lambda,\lambda'\in\Lambda, let P(k|\lambda,\lambda',X) denote the probability assigned by the ontological model for the two-qubit quantum theory to the result k, given that we're measuring X and that the ontic state of the two-qubit system is (\lambda,\lambda').
Now let \lambda be an ontic state of a single qubit that's assigned a non-zero probability by both Q_0 and Q_+. Define q=\min\{Q_0(\lambda),Q_+(\lambda)\}. Since |0\rangle\otimes|0\rangle is orthogonal to |\xi_1\rangle, we have
[tex]0=\left|\langle\xi_1| \left(|0\rangle\otimes| 0\rangle\right)\right|^2 =P_{00}(1|X)=\sum_{i=1}^n \sum_{j=1}^n Q_{00}(\lambda_i,\lambda_j) P(k|\lambda_i,\lambda_j,X)[/tex] Since every term is non-negative, this implies that all terms are 0. In particular, the term with \lambda_i=\lambda_j=\lambda is 0.
[tex]0=Q_{00}(\lambda,\lambda)P(1|\lambda,\lambda,X)[/tex]
Since Q_{00}(\lambda,\lambda) =Q_{0}(\lambda)Q_{0}(\lambda)\geq q^2>0, this implies that P(1|\lambda,\lambda,X)=0.
A very similar argument based on the fact that |0\rangle\otimes|+\rangle is orthogonal to |\xi_2\rangle implies that P(2|\lambda,\lambda,X)=0. A similar argument works for all four values of k, so we can prove that P(k|\lambda,\lambda,X)=0 for all k\in\{1,2,3,4\}. This implies that \sum_{k=1}^4 P(k|\lambda,\lambda,X)=0\neq 1. This implies that at least one of the assumptions that told us that we were dealing with a ψ-epistemic ontological model for the two-qubit quantum theory must be false.