Let me describe a 6-point counterexample.

Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do). Note that $K$ is CAT(0).

Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x_1,x_2,x_3,x_4,x_5\}$ of the rotation by angle $\tfrac\theta5$.

Suppose that these 6 points admit an embedding into a product of thees, say $L$. Let $q$ be the image of $p$; denote by $\Sigma_q$ its space of directions. Note that any closed geodesic in $\Sigma_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)

Denote by $y_i$ the image of $x_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy_{i-1}y_iy_{i+1}$ in $L$ that is isometric to the quadrilateral $px_{i-1}x_ix_{i+1}$. It follows that there is a closed geodesic in $\Sigma_q$ of length $\theta$ --- a contradiction.

Comments:

• There is a similar example with 5 points --- take a plane convex quadrilateral with line segment attached at one vertex. By Reshetnyak theorem, it is a CAT(0) space. The vertices of the quadraliteral + the end of the segment form a 5-point subspace that cannot be embedded in a product of trees.

• Another related observation: it is not hard to see that octahedron comparison described at the end of our bipolar comparison holds for products of trees. It is unknow if it holds in general CAT(0) space.

Let me describe a 6-point counterexample.

Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do). Note that $K$ is CAT(0).

Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x_1,x_2,x_3,x_4,x_5\}$ of the rotation by angle $\tfrac\theta5$.

Suppose that these 6 points admit an embedding into a product of thees, say $L$. Let $q$ be the image of $p$; denote by $\Sigma_q$ its space of directions. Note that any closed geodesic in $\Sigma_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)

Denote by $y_i$ the image of $x_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy_{i-1}y_iy_{i+1}$ in $L$ that is isometric to the quadrilateral $px_{i-1}x_ix_{i+1}$. It follows that there is a closed geodesic in $\Sigma_q$ of length $\theta$ --- a contradiction.

Comments:

• There is a similar example with 5 points --- take a plane convex quadrilateral with line segment attached at one vertex. By Reshetnyak theorem, it is a CAT(0) space. The vertices of the quadraliteral + the end of the segment form a 5-point subspace that cannot be embedded in a product of trees.

• Another related observation: octahedron comparison holds for products of trees. That is, if you choose six points in the product of trees and label them by vertices of an octahedron, then there is a configuration in the Euclidean space such that edges are not getting larger and the diagonals are not getting smaller. It is unknow if it holds in general CAT(0) space.

Let me describe a 6-point counterexample.

Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do). Note that $K$ is CAT(0).

Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x_1,x_2,x_3,x_4,x_5\}$ of the rotation by angle $\tfrac\theta5$.

Suppose that these 6 points admit an embedding into a product of thees, say $L$. Let $q$ be the image of $p$; denote by $\Sigma_q$ its space of directions. Note that any closed geodesic in $\Sigma_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)

Denote by $y_i$ the image of $x_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy_{i-1}y_iy_{i+1}$ in $L$ that is isometric to the quadrilateral $px_{i-1}x_ix_{i+1}$. It follows that there is a closed geodesic in $\Sigma_q$ of length $\theta$ --- a contradiction.

Comments:

• For 5-point spaces this is true. With some extra work it can be extracted from charcterization of 5-point sets in CAT(0) space, see the paper of Tetsu Toyoda or our shorter proof.

• Another related observation: it is not hard to see that octahedron comparison described at the end of our bipolar comparison holds for products of trees. It is unknow if it holds in general CAT(0) space.

Let me describe a 6-point counterexample.

Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do). Note that $K$ is CAT(0).

Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x_1,x_2,x_3,x_4,x_5\}$ of the rotation by angle $\tfrac\theta5$.

Suppose that these 6 points admit an embedding into a product of thees, say $L$. Let $q$ be the image of $p$; denote by $\Sigma_q$ its space of directions. Note that any closed geodesic in $\Sigma_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)

Denote by $y_i$ the image of $x_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy_{i-1}y_iy_{i+1}$ in $L$ that is isometric to the quadrilateral $px_{i-1}x_ix_{i+1}$. It follows that there is a closed geodesic in $\Sigma_q$ of length $\theta$ --- a contradiction.

Comments:

• There is a similar example with 5 points --- take a plane convex quadrilateral with line segment attached at one vertex. By Reshetnyak theorem, it is a CAT(0) space. The vertices of the quadraliteral + the end of the segment form a 5-point subspace that cannot be embedded in a product of trees.

• Another related observation: it is not hard to see that octahedron comparison described at the end of our bipolar comparison holds for products of trees. It is unknow if it holds in general CAT(0) space.