EDIT: As pointed out by Tapio Rajala, my proof is wrong without assuming that $X$ is compact. I've added this assumption, which one can drop by modifying the proof (I'll try to modify my answer to deal with the general case)

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Let $(X,d)$ denote a compact Polish length space and $(Geod(X),d_\infty)$ denote the (compact) Polish space of geodesics in $(X,d)$, equipped with the $\sup$-distance. Now, consider the map
$$
Eval: Geod(X) \to X\times X.
$$
which takes $\gamma\mapsto (\gamma(0),\gamma(1))$ (in this setting, it is standard that all geodesics are assumed to be of unit length, parametrized by constant speed).

<b>Claim 1</b>: The $Eval$ map is continuous. I'll leave this to you to check (its easy).

<b>Claim 2</b>: The $Eval$ map is surjective. This follows because we have assumed that $X$ is a length space, so there is a geodesic between any two points.

<b>Claim 3</b>: The $Eval$ map has compact fibers. This is clear because I've changed the assumption to $X$ closed.


Thus, we may apply measurable selection to $Eval$, per the version of measurable section on Villani (OT: Old and New) p. 92:

> A surjective Borel map between Polish spaces with compact fibers admits a Borel right inverse.


In particular, there exists
$$
GeodSel: X\times X \to Geod(X)
$$
so that $Eval(GeodSel(x,y)) = (x,y)$. This is exactly what you would like to use in the proof of the statement you mention in your question. 

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I think that the confusion between this question and the one you link to is that you are considering the multiple valued map $S$ from $X\times X$ to $2^{Geod(X)}$. This is your mistake: measurable selection constructs a _single valued_ such $S$. In order to construct such an $S$, you need to look at the evaluation map in the other direction, which _is_ a single valued map.