non-separable assignments of the vertices of a hypercube
You appear to be correct. I believe this is a simple proof of the opposite fact:

Let $s(n)$ denote the number of separable sets in the $n$-cube. Let $Q_0, Q_1$ denote the subcubes of one smaller dimension, with say last coordinates equal to 0, 1 respectively. The key fact is that any separable set of the whole cube intersects with each of $Q_0, Q_1$ in separable sets. (The separating hyperplane intersects the codimension-1 spaces to form separating hyperplanes.) So 
$$s(n)\leq [s(n-1)]^2.$$

Now the proportion $r(n)$ satisfies
$$r(n)=
\frac{s(n)}{2^{2^n}}\leq
\frac{[s(n-1)]^2}{(2^{2^{n-1}})^2}=
[r(n-1)]^2&lt;
r(n-1),$$
where the last inequality holds because $r(n-1)&lt;1$ for $n&gt;2$.