Question on geodesically convex set Let $(X,d)$ be a geodesic metric space (this means that every couple of points in $X$ are joined by a minimizing geodesic) and $C\subset X$ a geodesically convex set (given any couple of points in $C$ the minimizing geodesic joining them is entirely contained in $C$).
Is it true that for any point $x\in X\setminus C$ there is always a point $x'\in \partial C$ (the border of $C$) such that $d(x,c)\ge d(x',c)$ for every point $c\in C$?
 A: No. There is counterexample :
(1) Construction of space $X$ : Consider $B$, unit disc in
$\mathbb{R}^2$. If $C,\ D$ are isometric to $B$, then $X$ is obtained from identifying boundaries of $C,\ D$. Here
$C$ in $X$ is geodesically
convex. (That is intrinsic metric on $X$ is like a boundary surface of a coin)
(2) In $X$, there is $x\in X$ s.t. $x$ does not have $x'$ 
Proof : If $x$ is center of $D$, i.e. $d(x,\partial D)=1$, then assume that $x$ has $x'$ 
Consider a point $c_1\in C$, which is close
to $\partial C$, i.e. $$d(c_1,\partial C)=\epsilon$$
Hence $ d(x,c_1)=1+\epsilon$ so that $$d(x',c_1)\leq 1+\epsilon $$
(That is $x'$ can not escape from some hemisphere in $\partial C$.)
If $c_2\in C$ s.t. $d(c_1,c_2)=2-2\epsilon$ and a geodesic $[c_1c_2]$ contains
a center of $C$, then $ d(x,c_2)=1+\epsilon$.
So $d(x',c_2)$ is at least $d$, which is close to $\sqrt{3}$. Hence $d(x',c_2)>d(x,c_2)$. It is a contradiction.
Explanation about $\sqrt{3}$ : In $C\subset X$, for convenience, let $\epsilon=0$. Then
$d(c_1,c_2)=2$. Since $x'$ is around $c_1$ and $x'$ is in $\partial
C$, so $c_1c_2x'$ forms a right triangle. When $d(x',c_1)=1$,
$d(x',c_2)$ is smallest. 
