This is not possible. First lemma: There exist four points in $ A$ that are the vertices of a quadrilateral, convex or concave, which contains a fifth point in $ A$ inside it. That is, there exists some quadrilateral (whose vertices are in $ A$) that has another point in $ A$ inside it. Proof: Choose any five points in $ A$. Either these five satisfy the lemma, or the five points are the vertices of a convex pentagon. If the five points form a convex pentagon $ VWXYZ$, then each of the three disjoint triangles $ VWX$, $ VXY$, and $ VYZ$ must contain one point in $ B$. These three points in $ B$ form a triangle in which there must be a point in $ A$. But this new point in $ A$ is inside $ VWXYZ$! So we will be able to find four vertices of the pentagon which form a quadrilateral that contain this new point. Thus the lemma is proved. Second lemma: If a quadrilateral $ VWXY$ contains a point $ P$ in its interior as in the first lemma (All five points are in set $ A$), then there exists another point in set $ A$ that lies in the convex hull of the interior of the quadrilateral. Proof: If $ VWXY$ is concave with $ \angle WXY > \pi$, then the three disjoint triangles $ VWX$, $ VXY$, $ WXY$ each contain a point in $ B$. These three points in turn determine a triangle in which a new point in $ A$ lies. (If this point happens to be $ P$, then we just consider triangle $ PWX$ instead of $ VWX$, or triangle $ PXY$ instead of $ VXY$, depending on where $ P$ lies.) If $ VWXY$ is convex, then the three triangles $ VWP$, $ WXP$, $ XYP$ determine three points in $ B$ which in turn determine a new point in $ A$ in the interior. After applying the second lemma to some five points, we will be left with a quadrilateral with two points in its interior. We can now choose one of the points such that the remaining points satisfy lemma 1. The new quadrilateral will then be no larger than the first. wee can repeatedly aply lemma 2 and keep getting smaller quadrilaterals. But all of this must stop eventually, because there is a minimum distance between any two points! So this is impossible.

Lemma 1: Given any convex set of $5$ points in $A$, there is another point inside the points that is in $A$. Proof: Triangulate the set, we have enough points to form a triangle with points in $B$, thus there must be a another point in $A$. $\boxed{}$. Take $25$ points that belong to $A$. If we take the convex hull, by Lemma 1 there are at least five points inside this convex hull. Assume that are $a$ points in this convex hull that belong to $A$ (including the points on the convex hull) and let $b$ be the number of points in the convex hull that are in $B$. (Both of these numbers are finite since there is at most one point in any disc of diameter one.) Lemma 2: Given a set of $d$ points with $c$ points in the convex hull (not on the convex hull), there exists a triangulation of the $d$ points with $d-2+c$ triangles. Proof: If $c=0$, we do a usual triangulation. If $c=1$, draw a line from each point in the convex hull to the point. We can prove it for $c\ge 2$ by adding another point $P$ inside this triangulation and drawing lines from that $P$ to the verticies of the triangle that it is in. $\boxed{}$ Since there are at least $5$ points in $A$ in the convex hull (not counting the points on the convex hull), there are at least $(a-2)+5$ triangles in the triangulation of these $a$ points in $A$. Thus, $b\ge a+3$. If we triangulate the $b$ points in $B$, there are at least $b-2$ triangles. Thus, $a\ge b-2$. In conclusion, $b\ge a+3\ge b+1$, contradiction. edit: So no such sets of points exists.