EDIT: As pointed out by Tapio Rajala, my proof is wrong without assuming that $X$ is compact. I've added this assumption, which I don't think is necessary, but I am having some trouble seeing how to drop it.
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).
Claim 1: The $Eval$ map is continuous. I'll leave this to you to check (its easy).
Claim 2: 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.
Claim 3: 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.
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.