 Solution

Write $L=d(x,y)$. If $m^{\prime }$ is any other metric midpoint, apply the inequality from (a) to the chosen midpoint $m$ with $z=m^{\prime }$. The right side is $\frac{1}{2}(L/2{)}^2+\frac{1}{2}(L/2{)}^2-L^2/4=0$, so $m=m^{\prime }$. Every pair has a unique midpoint.

Choose successive midpoints to define $\gamma (k/2^n)$ on all dyadic parameters. Consecutive points at level $n$ have distance $L/2^n$. The triangle inequality gives $d(\gamma (s),\gamma (t))\le (t-s)L$ for dyadic $s<t$. Conversely,which gives the reverse bound. Thus equality holds. Completeness extends this map uniquely and continuously to $[0,1]$, retaining $d(\gamma (s),\gamma (t))=|s-t|L$. It is a constant-speed geodesic.

Every other geodesic has the same midpoint, and then the same successive midpoints, so agrees on all dyadic parameters. Continuity gives agreement everywhere. Hence $\boxed{\text{there is exactly one geodesic segment between any two points}.}$