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 [itex]\Lambda[/itex]. Then [itex]\Lambda\times\Lambda[/itex] 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 [itex]\Lambda[/itex] 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 [itex]\Lambda[/itex] by [itex]\lambda_1,\dots,\lambda_n[/itex].
Let [itex]\mathcal H[/itex] be the Hilbert space of the quantum theory of a single qubit. Then [itex]\mathcal H\otimes\mathcal H[/itex] is the Hilbert space of the quantum theory of two qubits. Let [itex]\{|0\rangle,|1\rangle\}[/itex] be an orthonormal basis for [itex]\mathcal H[/itex]. Define
[tex]
\begin{align}<br />
|+\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle+|1\rangle\right)\\<br />
|-\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle-|1\rangle\right).<br />
\end{align}[/tex] [itex]\{|+\rangle,|-\rangle\}[/itex] is another orthonormal basis for [itex]\mathcal H[/itex].
For each [itex]|\psi\rangle\in\mathcal H[/itex] and each [itex]\lambda\in\Lambda[/itex], let [itex]Q_\psi(\lambda)[/itex] denote the probability that the qubit's ontic state is [itex]\lambda[/itex]. The function [itex]Q_\psi:\Lambda\to[0,1][/itex] is called the epistemic state corresponding to [itex]|\psi\rangle[/itex]. Similarly, for each [itex]|\psi\rangle\otimes|\psi'\rangle\in\mathcal H\otimes\mathcal H[/itex] and each [itex](\lambda,\lambda')\in\Lambda\times\Lambda[/itex], let [itex]Q_{\psi\psi'}(\lambda,\lambda')[/itex] denote the probability that the two-qubit system is in ontic state [itex](\lambda,\lambda')[/itex]. 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 [itex]\mathcal H\otimes\mathcal H[/itex] with the eigenvectors
[tex]
\begin{align}<br />
|\xi_1\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle\otimes|1\rangle +|1\rangle\otimes|0\rangle\right)\\<br />
|\xi_2\rangle &=\frac{1}{\sqrt{2}} \left(|0\rangle\otimes|-\rangle +|1\rangle\otimes|+\rangle\right)\\<br />
|\xi_3\rangle &=\frac{1}{\sqrt{2}} \left(|+\rangle\otimes|1\rangle +|-\rangle\otimes|0\rangle\right)\\<br />
|\xi_4\rangle &=\frac{1}{\sqrt{2}} \left(|+\rangle\otimes|-\rangle +|-\rangle\otimes|+\rangle\right)<br />
\end{align}[/tex] Note that each of the state vectors
[tex]
\begin{align}<br />
&|0\rangle\otimes|0\rangle\\<br />
&|0\rangle\otimes|+\rangle\\<br />
&|+\rangle\otimes|0\rangle\\<br />
&|+\rangle\otimes|+\rangle<br />
\end{align}[/tex] is orthogonal to exactly one of the [itex]|\xi_k\rangle[/itex].
The result of an X measurement that corresponds to eigenvector [itex]|\xi_k\rangle[/itex] will be denoted by [itex]k[/itex]. For all [itex]k[/itex] and all [itex]\psi,\psi'\in\mathcal H[/itex], let [itex]P_{\psi\psi'}(k|X)[/itex] 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 [itex]Q_{\psi\psi'}[/itex]. For each [itex]k[/itex] and each [itex]\lambda,\lambda'\in\Lambda[/itex], let [itex]P(k|\lambda,\lambda',X)[/itex] 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 [itex](\lambda,\lambda')[/itex].
Now let [itex]\lambda[/itex] be an ontic state of a single qubit that's assigned a non-zero probability by both [itex]Q_0[/itex] and [itex]Q_+[/itex]. Define [itex]q=\min\{Q_0(\lambda),Q_+(\lambda)\}[/itex]. Since [itex]|0\rangle\otimes|0\rangle[/itex] is orthogonal to [itex]|\xi_1\rangle[/itex], 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 [itex]\lambda_i=\lambda_j=\lambda[/itex] is 0.
[tex]0=Q_{00}(\lambda,\lambda)P(1|\lambda,\lambda,X)[/tex]
Since [itex]Q_{00}(\lambda,\lambda) =Q_{0}(\lambda)Q_{0}(\lambda)\geq q^2>0[/itex], this implies that [itex]P(1|\lambda,\lambda,X)=0[/itex].
A very similar argument based on the fact that [itex]|0\rangle\otimes|+\rangle[/itex] is orthogonal to [itex]|\xi_2\rangle[/itex] implies that [itex]P(2|\lambda,\lambda,X)=0[/itex]. A similar argument works for all four values of [itex]k[/itex], so we can prove that [itex]P(k|\lambda,\lambda,X)=0[/itex] for all [itex]k\in\{1,2,3,4\}[/itex]. This implies that [itex]\sum_{k=1}^4 P(k|\lambda,\lambda,X)=0\neq 1[/itex]. 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.