What is the minimum number of colors that always suffice to color every planar set of points such that any disk that contains enough points contains two points of different colors? It is known that the answer to this question is either three or four.We show that three colors always suffice if the condition must be satisfied only by disks that contain a fixed point.Our result also holds, and is even tight, when instead of disks we consider their topological generalization, namely \emph{pseudo-disks}, with a non-empty intersection.Our solution uses the equivalence that a hypergraph can be realized by stabbed pseudo-disks if and only ifit is \emph{$ABAB$-free}.These hypergraphs are defined in a purely abstract, combinatorial way and our proof that they are $3$-chromatic is also combinatorial.