Is the Pusey-Barrett-Rudolph (PBR) theorem only applicable in two dimensions? In the Pusey-Barrett-Rudolph (PBR) paper, https://arxiv.org/abs/1111.3328 it is shown that two non-colinear  pure states $|\psi_0\rangle$ and $|\psi_1\rangle$ cannot have overlapping ontic supports (modulo some assumptions). However, the demonstration given in the paper seems to be done only for a two dimensional Hilbert space.
Does the strength of the PBR argument cover quantum states that are elements of arbitrary dimensional Hilbert spaces?
 A: Yes, the argument given in that paper is completely general and applies to pairs of states in Hilbert spaces of arbitrary dimension. The choose two linearly independent vectors and look at the two-dimensional subspace generated by these vectors. They do not assume that the Hilbert space itself is two-dimensional. In fact, they they say this explicitly:

These states span a two-dimensional subspace of the Hilbert space. We can restrict attention to this subspace and from hereon, without loss of generality, treat the systems as qubits.

This is a perfectly standard approach to theorems in linear algebra that is used all the time. We use it in QM very often too, e.g. when we diagonalize operators simultaneously, where we fix the eigenvalue for one of them (i.e., we restrict the attention to one of its eigenspaces) and study the eigenvalues of the other one.
A: Motivated by AccidentalFourierTransform's answer I decided to fill in some of the details.
Let $|\psi_0\rangle$ and $|\psi_1\rangle$ be two normalized non-colinear elements of an $N$ dimensional Hilbert space $\mathcal{H}$.  They then span a two dimensional subspace $\mathcal{S}_2$ of $\mathcal{H}$.  A basis $|0\rangle$, $|1\rangle$,...,$|N-1\rangle$ for $\mathcal{H}$ exists such that $|0\rangle$ and $|1\rangle$ span $\mathcal{S}_2$.
Following the first example given in the PBR paper, let
\begin{equation}
     |\psi_0\rangle=|0\rangle,\:\:\:\:\:|\psi_1\rangle=|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle).
\end{equation}
Then, as discussed in the paper, bringing two copies of the these quantum states together generates an $N^2$ dimensional Hilbert space $\mathcal{H}\otimes\mathcal{H}$, which in particular has the four elements $|0\rangle\otimes|0\rangle$, $|0\rangle\otimes|+\rangle$, $|+\rangle\otimes|0\rangle$, and  $|+\rangle\otimes|+\rangle$, each of which is compatible with the physical state $\lambda$. These four vectors as well as $|\xi_1\rangle$, $|\xi_2\rangle$, $|\xi_3\rangle$, and $|\xi_4\rangle$ (as defined in the paper) span the same four dimensional subspace $\mathcal{S}_4$ of $\mathcal{H}\otimes\mathcal{H}$. Since the $|\xi_i\rangle$ elements are orthonormal, an additional $N^2-4$ orthonormal elements exist so that $\{|\xi_i\rangle\}$ can be extended into a orthonormal basis for $\mathcal{H}\otimes\mathcal{H}$. Then there exists a projective measurement onto this basis. As stated in the paper, $\langle\xi_1|(|0\rangle\otimes|0\rangle)=0$, $\langle\xi_2|(|0\rangle\otimes|+\rangle)=0$, $\langle\xi_3|(|+\rangle\otimes|0\rangle)=0$, $\langle\xi_4|(|+\rangle\otimes|+\rangle)=0$. But, additionally each of these is also orthogonal to the remaining $|\xi_i\rangle_{i>4}$, since each $|\xi_i\rangle_{i>4}$ is orthogonal to any element of $\mathcal{S}_4$. Thus if the physical state $\lambda$ happens to be prepared and a $\hat{\xi}$ measurement is made, every possible outcome has zero probability of being realized, which contradicts the tenets of QM.
A similar argument can be applied to the next part of the paper where two arbitrary normalized non-colinear elements  $|\psi_0\rangle$ and $|\psi_1\rangle$ are considered. As discussed in the paper n copies of these states are brought together generating a $N^n$ dimensional Hilbert space $\mathcal{H}^{\otimes{n}}$. This Hilbert space contains the elements $|\psi_{x_1}\rangle\otimes...\otimes|\psi_{x_n}\rangle$ where $x_i\in\{0,1\}$, which taken together span a $2^n$ dimensional subspace $\mathcal{S}_{2^n}$ of $\mathcal{H}^{\otimes{n}}$. The authors also provide an orthonormal basis $\{(H^{\otimes{n}}R_{\alpha}Z_{\beta}^{\otimes{n}})^{\dagger}|x_1...x_n\rangle\}$ for $\mathcal{S}_{2^n}$ such that for each $|\psi_{x_1}\rangle\otimes...\otimes|\psi_{x_n}\rangle\in\mathcal{S}_{2^n}$
\begin{equation} 
\langle{x_1...x_n}|{H}^{\otimes{n}}R_{\alpha}Z_{\beta}^{\otimes{n}}|\psi_{x_1}\rangle\otimes...\otimes|\psi_{x_n}\rangle=0.
\end{equation}
As in the previous example $\{(H^{\otimes{n}}R_{\alpha}Z_{\beta}^{\otimes{n}})^{\dagger}|x_1...x_n\rangle\}$ can be extended into an orthonormal basis for $\mathcal{H}^{\otimes{n}}$ such that each basis element is orthogonal to at least one $|\psi_{x_1}\rangle\otimes...\otimes|\psi_{x_n}\rangle$. Assuming again that the physical state $\lambda$ is compatible with each $|\psi_{x_1}\rangle\otimes...\otimes|\psi_{x_n}\rangle$, then a measurement projecting onto the aforementioned basis will have no realizable outcome.
