Let $X$ be a convex subset of a separable Banach space and $A$ a nonempty subset of $X$ with the following property: for any point $a\notin A$ and any $x\in X$, there are finitely many elements of $A$ contained in the segment $[a,x]=\{\lambda a+(1-\lambda)x: \lambda\in [0,1]\}$. Does it follow that $A$ is nowhere dense in $X$? I thought the solution might have something to do with the Baire category theorem, but it's difficult to account for the uncountability of segments.
EDIT: The property implies that the complement $X\setminus A$ is dense, but nowhere denseness of $A$ is a stronger assertion. This is not a homework problem but came up in my research where $X$ is the space of measures on a given domain.
EDIT 2: The question was originally posed for a fixed $a\notin A$ as well, but I have since come up with a counterexample. Let $X=\mathbb{R}^2$, $A=\mathbb{Q}^2$ the subset of all points with rational coordinates and $a=(\sqrt{2},\sqrt{3})$. Then any straight line $y=mx+b$ passing through $a$ cannot intersect with $A$ at two or more points, since that would imply $m,b\in\mathbb{Q}$ but $\sqrt{2},\sqrt{3}$ are rationally independent. However, this idea does not extend to all $a\notin A$, for example the segment connecting $(\sqrt{2},0)$ and $(0,0)$ contains infinitely many points in $A$.
Let $X$ be any second-countable topological vector space of dimension greater than $1$; I will construct a dense subset $A\subset X$ that satisfies your property. Let $(U_n)$ be a countable basis for the topology of $X$. Recursively choose a sequence of points $(a_n)$ such that $a_n\in U_n$ and $a_n$ is not on the line between any two $a_i,a_j$ for $i,j<n$ for each $n$ (this is possible since $\dim X>1$, so these finitely many lines cannot cover $U_n$). Then $A=\{a_n:n\in\mathbb{N}\}$ is dense in $X$, and has the property that no three points of $A$ are collinear, so in particular no line segment contains infinitely many points of $A$.