Parallel planes of minimal surface If $S_{1}$ and $S_{2}$ are properly embedded minimal surfaces with free boundary set. If $S_{1}$ does not intersect $S_{2}$ and there exist $p_{1} \in S_{1}$ and  $p_{2} \in S_{2}$ such that $dist(S_{1}, S_{2})=dist(p_{1}, p_{2})$.
How we can prove that $S_{1}$ and $S_{2}$ are parallel planes?
I can prove that if $S_{1}$ and $S_{2}$ are proper, possibly branched minimal surface depending on Strong half space theorem and Hoffman- Meeks theorem but for the embedded and free boundary I am totaly lost.
Any advise please.
 A: Outline:
I do assume you are looking at the ambient space $\mathbb{R}^3$ and the $p_i$ are interior points of the minimal surfaces. I also do assume you know minimal surfaces cannot touch in an interior point with one of them lying locally completely on one side of the other one, unless they coincide.
(By saying a regular surface $M_1$ is locally lying completely on one side of another regular $M_2$ near $p\in M_2$ I mean that if $M_2$ subdivides a sufficiently small ball around $p$ in two connected open components, then the intersection of $M_1$ with one of these open connected components is empty).
If there are two points (interior with respect to the respective surface) which minimize the distance, then the line joining these two points will be orthogonal to both surfaces (why?) in the $p_i$, in other words, the tangent planes of $\cal S_i$ in $p_i$ are parallel. This means you can, in a neighbourhood of $p_1$, say, write both $S_1$ and $S_2$ as a graph over the tangent plane $T_{p_1}\cal S_1$. Since the distance minimum is realized in $p_1, p_2$, you can now (again in a neighbourhood of $p_1$) translate $S_2$ along the normal to $T_{p_1}\cal S_1$ in direction $S_1$ until the two surfaces touch for the first time. This happens necessarily in $p_1$, by assumption. But then, locally, they touch in one single point while each of them is lying completely on side of the other one -- in contradiction to the statement I made in the beginning.
Edit (thanks to davidivadful): This only shows that the two surfaces are translates of each other, they need not be planes. If you want to derive that, as well, you need to have additional Information, e.g. about the boundary.
