Suppose we have a normed plane. If it helps, assume the norm is strictly convex. Take your favourite point on the unit sphere $(X,Y)$ and take two points $(a,b)$ and $(c,d)$ where the unit sphere at $(X,Y)$ meets the unit sphere at the origin:
This gives a quadrilateral with vertices at $(X,Y)$, $(a,b)$, $(c,d)$, and the origin.
The diagonal from the origin to $(X,Y)$ has length one unit by construction. I am interested in the other diagonal. It seems that the diagonal from $(a,b)$ to $(c,d)$ has length at least one. In fact it seems every point on the (red) side from $(X,Y)$ to $(a,b)$ has distance one or greater from $(c,d)$ and likewise for the blue side from $(X,Y)$ to $(c,d)$. In other words the red unit ball intersects the blue side at some segment that contains $(X,Y)$.
The above image is made using Desmos. The above statement is true for every norm and choice of points I have tried. Feel free to drag the points around yourself to convince yourself. The above norm is the $p=1.8$ norm. The linked toy can generate figures for other $p$-norms too.
In any case, this seems like an inequality that, if true, should be provable using only the basic norm axioms and triangle inequality. But it's just not clicking for me at the moment. Is it true? How come?