Is a lattice of convex sets in $R^2$ distributive?

In the paper "On nonmodular ndistributive lattices: Lattices of convex sets" (Acta Sci. Math. (Szeged) 52 (1987), 3545) A. Hunh proves that the lattice of convex sets in $\mathbb R^n$, $Co(\mathbb R ^n)$, is $n+1$distributive but not $n$distributive. This means that the lattice of convex sets in $\mathbb R^n $ is only distributive for $n=0$, as the example by François G. Dorais shows. Here we define a lattice $L$ to be $n$distributive if for any $x,y_0,\dots,y_n$ the following identity holds: $$x\wedge \bigvee_{i=0}^n y_i =\bigvee_{i=0}^n (x \wedge \bigvee_{i\neq j} y_i).$$ This nice result shows that the dimension of a Euclidean space has a lattice theoretical characterization. The proof relies on Caratheodory's theorem. The same holds for the dual of $Co(\mathbb{R}^n)$, it is $n+1$distributive but not $n$distributive, and the proof uses Helly's theorem. 


I googled "lattice of convex sets"+distributive and the second link led me to the abstract for the paper: "Geometric Condition for Local Finiteness of a Lattice of Convex Sets" in Mathematica Moravica, Vol. 1 (1997), 35–40 by Matt Insall, which contains the following sentence:
This means that the answer to your question is no, since $\mathbb{R}^2$ is certainly a Hilbert space. Unfortunately, the journal did not have the paper online, but I was able to find Insall's webpage and from there, the PDF for the paper. The paper is relatively easy to read and has illustrations for the examples. 

