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)

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.

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.

Let $(X,d)$ denote a Polish length space and $(Geod(X),d_\infty)$ denote the 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. Geometrically, this is the statement that the set of geodesics between $x$ and $y$ is compact with respect to $d_\infty$. This follows from Arzelà–Ascoli (constant speed geodesics are always equicontinuous) and the triangle inequality shows that they are always "equibounded" i.e., given a fixed geodesic $\gamma$ between $x$ and $y$, any other geodesic $\gamma'$ satisfies $d_\infty(\gamma,\gamma') \leq C$ for some $C$. To check this, note that if it failed, we could find $\gamma'$ with $d_\infty(\gamma,\gamma') > 100d(x,y)$. Then, there is some point $z \in \gamma'$ which is at least $100d(x,y)$ away from $\gamma$. Thus, it is at least $100d(x,y)$ away from $x$ and $y$. Thus, $length(\gamma') > 100d(x,y)$, which contradicts it being a geodesic.

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.

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)

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.