A subset of a geodesic metric space is called convex if for every two points in the subset one of the geodesics connecting these points lies in the subset. Is it true that every convex subset of a product of two trees with $l_1$metric is a median space, that is for every three points A,B,C in the subset there exists a point D in the subset such that D is on (some) geodesics connecting A and B, B and C, A and C?

Yes. Let $A=(A_1,A_2)$, $B=(B_1,B_2)$ and $C=(C_1,C_2)$. For the triangle $A_1B_1C_1$ in the first tree, there is a "center" $M_1$ such that the (unique) geodesics $[A_1B_1]$, $[A_1C_1]$ and $[B_1C_1]$ contain $M_1$. Similarly, there is a "center" $M_2$ for the triangle $A_2B_2C_2$ in the second tree. Let us show that $(M_1,M_2)$ belongs to any convex set $S$ containing $A$, $B$ and $C$. Suppose $A_1\ne M_1$ and $A_2\ne M_2$. Then consider a geodesic $A(t)=(A_1(t),A_2(t))$ connecting $A$ to $B$ in $S$. Its coordinate projections are geodesics in the two trees. Let $A_1(t)$ hit $M_1$ before $A_2(t)$ hits $M_2$ and this first hit happens at $t=t_0$. Replace $A$ by $A'=A(t_0)$. It suffices to prove the assertion for $A'BC$ in place of $ABC$ (note that the centers of the new coordinate triangles are the same points $M_1$ and $M_2$). What we gained is that now one of $A_1$ and $A_2$ coincides with the corresponding center $M_1$ or $M_2$. Repeat this procedure for $B$ and $C$. Now, for each of $A$, $B$ and $C$, one of the coordinate projections is at $M_1$ or $M_2$. So at least two of these projections are at the same point $M_1$ or $M_2$. Assume w.l.o.g. that $A_1=B_1=M_1$. Consider a geodesic from $A$ to $B$ in $S$. Its first coordinate projection is constant $M_1$, and the second one is a geodesic from $A_2$ to $B_2$ that must go through $M_2$ eventually. 


I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take $\mathbb{R}^3$ with the $l^1$ metric and inside it the plane $x+y+z=1$. It is convex (because Euclidean segments are geodesics also for $l^1$), not strongly convex, and not median: it contains the points $(1,0,0), (0,1,0), (0,0,1)$ but not their median point $(0,0,0)$. 

