Convexity properties of subset of the plane in which among any n+1 points, one pair can be connected by line segment Say that a subset $S$ of $\mathbb R^2$ is $n$-solitary if, for any subset $P\subseteq S$ of size at least $n+1$, there exists at least one pair of distinct elements $p_1,p_2\in P$ such that the segment from $p_1$ to $p_2$ is wholly contained in $S$.

Is an $n$-solitary set necessarily the union of $n$ convex sets?


This question is inspired by this one on MathOverflow, which asks if $\mathbb R^2$ minus $n$ disjoint line segments is $n+1$-solitary.
Note that, clearly the union of $n$ convex sets satisfies this property by the pigeonhole principle. Obviously, being $1$-solitary is the same as being convex. I do not know any concrete results beyond this.
If we define $N_S(p)$ to be the set of points in $S$ that can be connected to $p$ by a line segment entirely within $S$. It seems productive to study the sets of the form $S\setminus N_S(p)$, but it is not clear how. For instance, for $n=2$ and any $p\in S$, we have that any pair of points in $S\setminus N_S(p)$ must be connected by a straight line.
Note that this is equivalent to the following more combinatorial problem:

Let the graph $G_S$ have vertex set $S$ and an edge $s_1$ to $s_2$ if there is a line segment connecting $s_1$ to $s_2$ within $S$. Does it follow that, if $G_S$ has no independent sets of size $n+1$, then it can be covered by $n$ cliques?

This statement is not true of general graphs $G$ (for instance, a cycle graph with $5$ vertices has no independent set of size $3$, but certainly has no clique-cover of size $2$), but it's not obvious how to characterize the set of graphs of the form $G_S$.
 A: A set $S\subseteq\mathbb R^d$ is called $m$-convex if every set of $m$ points in $S$ contains a pair of points such that the line segment joining them lies entirely in $S;$ thus $n$-solitary $=$ $(n+1)$-convex.
Quoting the abstract of "A planar $3$-convex set is indeed a union of six convex sets" by Noa Nitzan and Micha A. Perles:

Suppose $S$ is a planar set. Two points $a, b$ in $S$ see each other via $S$ if $[a, b]$ is included in $S$ . F. Valentine proved in 1957 that if $S$ is
  closed, and if for every three points of $S$ , at least two see each other
  via $S$ , then $S$ is a union of three convex sets. The pentagonal star
  shows that the number three is best possible. We drop the condition
  that $S$ is closed and show that $S$ is a union of (at most) six convex
  sets. The number six is best possible.

Nitzan and Perles give the following references:

Breen, M. (1974), Decomposition theorems for $3$-convex subsets of the plane,
  Pacific J. Math., Vol. 53, 43-57.
Breen, M. and D. Kay (1976), General decomposition theorems for $m$-convex
  sets in the plane, Israel J. Math., Vol. 24, 217-233.
Lawrence J.F., W.R. Hare and J.W. Kenelly (1972), Finite unions of convex
  sets , Proc. Amer. Math. Soc., 34, 225-228.
Matoušek J. and P. Valtr (1999), On visibility and covering by convex sets,
  Israel J. Math., Vol. 113, 341-379.
Perles, M.A. and S. Shelah (1990), A closed $(n+1)$-convex set in $\mathbb R^2$ is a union of $n^6$ convex sets, Israel J. Math., Vol. 70, 305-312.
Valentine, F.A. (1957), A three point convexity property, Pacific J. Math.,
  7(2), 1227-1235.

See section A38 (pp. 46–47) of Unsolved Problems in Geometry by Hallard T. Croft, Kenneth J. Falconer, and Richard K. Guy for additional references.
