Imagine orienting the octahedron so that the two opposite faces are horizontal. Project onto a horizontal plane; these two faces are congruent equilateral triangles which (when projected) have the same center and opposite orientations. Hence, the vertices of the octahedron project to the vertices of a regular hexagon $A B C D E F$. Let $O$ be the center of the hexagon and $M$ the midpoint of $A C$. Now $A B M$ is a 30-60-90 triangle, so $A B=A M /(\sqrt{3} / 2)=(1 / 2) /(\sqrt{3} / 2)=\sqrt{3} / 3$. If we let $d$ denote the desired vertical distance between the opposite faces (which project to $A C E$ and $B D F)$, then by the Pythagorean Theorem, $A B^{2}+d^{2}=1^{2}$, so $d=\sqrt{1-A B^{2}}=\sqrt{6} / 3$.