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For $v,w$ in a real normed space, start with the points at distance $\Vert v-w\Vert /2$ from both endpoints and repeatedly retain those within half the previous diameter of every retained point. Reflection about $(v+w)/2$ preserves every stage, the midpoint belongs to every stage, and the diameters decrease by a factor at least two. Their intersection is therefore exactly the midpoint.

LetThe map $R$ is reflection about $m$ and is an isometry. It interchanges $v,w$, so it preserves $S_1^{vw}$. Inductively, if it preserves $S_{n-1}^{vw}$, then it preserves its diameter and all distances in the defining condition for $S_n^{vw}$; hence it preserves every $S_n^{vw}$.

The midpoint belongs to $S_1^{vw}$ becauseSuppose $m\in S_{n-1}^{vw}$. For every $z\in S_{n-1}^{vw}$, reflection invariance gives $R(z)\in S_{n-1}^{vw}$ andThus $m\in S_n^{vw}$, so all the sets are nonempty and contain $m$.

Put $D_n=\mathrm{diam}(S_n^{vw})$. If $x,y\in S_n^{vw}$ with $n\ge 2$, then $y\in S_{n-1}^{vw}$ and the defining property of $x$ givesThereforeIf $z$ lies in every $S_n^{vw}$, then both $z$ and $m$ lie in $S_n^{vw}$, soHence $z=m$. We have proved the metric extraction of a midpoint by shrinking diameters:

Give $S^2\times \mathbb{R}$ the Riemannian product of the unit round metric and the Euclidean metric. It is complete and has infinite diameter. The round sphere has scalar curvature $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, soThus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the Bonnet-Myers theorem. Its Ricci curvature vanishes in the $\mathbb{R}$ direction, showing precisely why a scalar-curvature bound is insufficient.