Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous.

After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur:

1. which metric spaces are central?
2. for which of the central spaces, is the central point operation $\ s\ $ continuous ?

Let me focus here on the second question, and this will be THE question in this thread:

QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space?

Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$:

First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$

Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central).

An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point.

(I'd be willing to provide more material).