Let $(X,d)$ be a complete metric linear space whose balls are convex. Let $Y\subseteq X$ be a bounded, closed and convex subset that verifies the following property: for all $y_0\in Y$, the distance function $y\rightarrow d(y_0,y)$ attains its sup in $Y$. Does $Y$ have extreme points?
I was trying to adapt the classical Krein-Milman's proof: the family of non-empty closed faces is not empty (because, given $y_0$, the set of maxima of $d(y_0,\cdot)$ is an extremal non-empty set and one can get a face by taking a maximal and convex superset of $\bar y$, being $\bar y$ a point that maximizes the distance from $y_0$). It is inductive by completeness, boundedness and convexity of the faces. The last step is the hardest one and I'm not sure it is true: given a face $F$ with two different points $x$ and $y$, can I say that $d(x,\cdot)$ attains its sup in $F$?
Or.. are there counterexamples?