I have a set $S\subset\mathbb {R}^2$ with the following property
(P) $\forall x,y\in S$, $\forall\mathscr{C}$ a convex set that contains $x$ in its interior, $bd\mathscr{C}\cap [x,y]\subset \overline{bd\mathscr{C}\cap S}$.
Here $[x,y]$ denotes the segment with end-ponts $x$, $y$, $bd\mathscr{C}$ denotes the boundary of $\mathscr{C}$, and "$\overline{\ \ \ \ \ }$" stands for closure.
In other words the inclusion says
$\forall z\in bd\mathscr{C},\ z=tx+(1-t)y$, for some $0<t<1$, there is $(z_n)_n\subset bd\mathscr{C}\cap S,\ z_n\to z$.
This feels like $S$ is dense in an orderly fashion in the segment $[x,y]$ (on every boundary of a convex set).
Conjecture: If $S$ has (P) then $S$ is pathwise connected (meaning there is a continuous path within $S$ that connects any two points in $S$).
My try: For fixed $x,y\in S$ define the multi-function $F:[0,1]\rightrightarrows\mathbb{R}^2$, $$F(t):=\cup\{bd\mathscr{C}\cap S\mid tx+(1-t)y\in bd\mathscr{C}\ {\rm and}\ \mathscr{C}\ {\rm is\ a\ convex\ set\ that\ contains\ } x \ {\rm in\ its\ interior}\}$$
If one could use a continuous selection theorem to get a continuous selection of $F$ then it would be done. Unfortunately, I cannot show that, for example, $F$ has convex values to use Michael's selection theorem.
P.S. Of course evey locally pathwise connected set of $\mathbb {R}^2$ has (P). The current question is exactly the converse of that fact.
Any remark is greatly appreciated.
A Bernstein set $B \subseteq \mathbb{R}^2$ should be a counterexample. Being Bernstein means both $B$ and $\mathbb{R}^2 \backslash B$ meet every perfect subset of $\mathbb{R}^2$.