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.

It meant to be an answer to another question, https://mathoverflow.net/a/406833/, but it is an answer to this question as well.