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? 



Existence of extreme points
There is a counter-example.


Note that in any normed space, its unit ball satisfies your supremum-attaining property. Indeed, for any $x\_0\in X$ the supremum of $d(x\_0,\cdot)$ on the ball is attained at the point $-x\_0/\|x\_0\|$ if $x\_0\ne 0$, and at any point of the sphere if $x\_0=0$.


It remains to construct a Banach space whose unit ball has no extreme points. Let $X$ be the space of all sequences $(x\_i)\_{i=1}^\infty$ such that $x\_i\to 0$ as $i\to\infty$, equipped with the $\ell\_\infty$ norm $\|x\|=\sup|x\_i|$. It is a closed subspace of $\ell\_\infty$ and hence complete. But clearly its unit ball has no extreme points.


