Examples of metric spaces with measurable midpoints
We will use the Kuratowski–Ryll-Nardzweski selection theorem:
Let $(\Omega, \mathscr{F})$ be a measurable space. Let $E$ be a Polish space. Let $\Gamma$ be a set-valued function from $\Omega$ to $E$; that is, for each $\omega \in \Omega$, let a set $\Gamma(\omega) \subseteq E$ be given. Assume that, for all $\omega \in \Omega$, the set $\Gamma(\omega)$ is nonempyty and closed in $E$. Assume that $\Gamma$ is $\mathscr F$-measurable in the sense:
$$
\text{for every open set }U\subseteq E,\qquad
\{\omega\,:\,\Gamma(\omega) \cap U \ne \varnothing\} \in \mathscr F .
$$
Then there is a measurable selection $\gamma$ for $\Gamma$: that is, a
function $\gamma : \Omega \to E$ with
$\bullet $ $\gamma(\omega) \in \Gamma(\omega)$
$\bullet $ for every open set $U \subseteq E,\quad \gamma^{-1}(U) \in \mathscr F$.
Let $X$ be a locally compact complete separable metric space with the midpoint property. For $a,b \in X$, let $\Gamma(a,b)$ be the midpoint set, $$ \Gamma(a,b) = \left\{m : d(a,m)=d(b,m) = \frac{d(a,b)}{2}\right\} . $$ Then $\Gamma$ is a set-valued function from $X \times X$ to $X$. Note $\Gamma(a,b)$ is nonempty and closed. Let $\mathscr F$ be the sigma-algebra of Borel sets in $X \times X$. We will prove (see below) that $\Gamma$ is $\mathscr F$-measurable. An application of the Kuratowski–Ryll-Nardzweski selection theorem then establishes the existence of an $\mathscr F$-measurable $\gamma : X\times X \to X$ with $\gamma(a,b) \in \Gamma(a,b)$.
Proof that $\Gamma$ is $\mathscr F$-measurable:
Let $U \subseteq X$ be open. We have to show $T_U \in \mathscr F$, where $$ T_U := \{(a,b) \in X \times X\,:\,\Gamma(a,b) \cap U \ne \varnothing\} . $$ The set $$ Q := \{(a,b,u) \in X \times X \times X \,:\, d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and } d(b,u) = \textstyle\frac{1}{2}d(a,b)\} $$ is a closed set. Write $\pi$ for the continuous "projection" function $(x,y,u) \mapsto (x,y)$. Then $T_U$ is the projection $$ T_U = \pi(Q\cap(X \times X \times U)) = \bigcup_{u \in U}\{(a,b) \in X \times X \,:\, d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and } d(b,u) = \textstyle\frac{1}{2}d(a,b)\} . $$ Now in our case, any open set $U$ is a countable union of compact sets, so the projection is sigma-compact, and therefore Borel.
added
Without assuming locally compact, we do know that the projection of
a Borel set is analytic and thus universally measurable. So if we are given a Borel measure $\mu$ on $X \times X$, we get a $\mu$-measurable midpoint function.
If $(X,d)$ is a complete metric space with the algebraic midpoint property (i.e. for all $x$ and $y$ in $X$, there exists $z\in X$ such that $d(x,z)=d(y,z)=d(x,y)/2$) then $X$ is a path metric space. Indeed, for all $x,y\in X$ one can iteratively construct a map $\gamma$ from $[0,1]\cap\mathbb D$ to $X$ such that $d(\gamma(s),\gamma(t))=|t-s|$, and extend it using completeness of $X$. Here $\mathbb D$ is the set of dyadic numbers. This result is Theorem 1.8 in [G].
Suppose that $(X,d)$ is complete, locally compact, and has the algebraic midpoint property. Then $X$ has the measurable midpoint property.
Of course, this does not include general closed convex subsets of Banach spaces, but it covers for instance any complete manifold.
Let us $(U_k)_{k\geq0}$ construct a countable basis of $X$; in particular, it will show that $X$ is second countable. I want the diameter of $U_k$ to tend to zero as $k$ goes to infinity, and every fixed $x$ to be contained in an infinite number of $U_k$. By [G, Theorem 1.10 (Hopf-Rinow)], the closed metric balls of $X$ are compact; then one can take a finite open cover of $B(x_0,1)$ by open balls of radius $1/1$, then a finite open cover of $B(x_0,2)$ by open balls of radius $1/2$, etc. Let also $(z_k)_{k\geq0}$ be a sequence such that $z_k\in U_k$.
Note that for any closed set $F$, the set of pairs $(x,y)$ such that $F$ contains at least one midpoint of $\lbrace x,y\rbrace$ is closed, using the compacity of closed bounded sets. Let $k_0(x,y)$ be the first $k$ such that the closure $\overline U_k$ contains at least one midpoint of $\lbrace x,y\rbrace$, and iteratively $k_{n+1}(x,y)$ is the first $k>k_n(x,y)$ such that the closed intersection $$ \overline U_k\cap\bigcap_{0\leq m\leq n}\overline U_{k_m(x,y)} $$ contains at least one midpoint of $\lbrace x,y\rbrace$.
Note that $k_n(x,y)$ is measurable, since the set of $(x,y)$ such that $k_n(x,y)\leq K$ is a finite union of closed sets. Then obviously $f_n:(x,y)\mapsto z_{k_n(x,y)}$ is measurable as well. Since the diameter of $U_k$ tends to zero, $(f_n(x,y))_{n\geq0}$ is a Cauchy sequence for all $(x,y)$, and $f(x,y):=\lim_{n\to\infty}f_n(x,y)$ is a well-defined midpoint of $\lbrace x,y\rbrace$. As a limit of measurable functions, it is measurable as well.
[G] M. Gromov, Metric structures for Riemannian and non-Riemannian spaces. 3rd printing (2007).
Maybe the unit circle embedded in the euclidean plane is an example of a space that has several measurable midpoints structures but no continuous such structure?
Let us choose as the middle point of two points which are not on a diameter the point in the middle of the shortest arc connecting the two points.
When two points are on a diameter, we may choose one of the middle points on the two arcs as our middle point, but this cannot be made continuously.