Let $(Z,d)$ be a Polish space, and for $p\geq 1$, consider a metric space $(W_p,d_{W^p})$ defined by
The Wasserstein Space $\begin{align}W_p = \{\mu|\mu\textrm{ is a Borel probability measure on Z such that} \int_{Z}d(z_0,z)^p\mu(dz)<\infty \textrm{ for some }z_0 \} \end{align}$
and
The Wasserstein distance $\begin{align} d_{W^p}(\mu,\nu) = \inf_{\pi}\left(\int_{Z\times Z}d(z,z')^p\pi (dz\times dz')\right)^{1/p}\end{align}$
where $\pi$ is a coupling between $\mu$ and $\nu$ i.e. a probability measure on $Z\times Z$ such that $\pi(A\times Z)=\mu(A), \pi(Z\times B)=\nu(B)$ for any measurable $A,B\subset Z$.
Now suppose that $(W_p,d_{W^p})$ is a geodesic space i.e. for any Borel probability measures $\mu,\nu$, there is a curve $\gamma:[0,1]\to W_p$ s.t. $\gamma(0)=\mu,\gamma(1)=\nu$ and $d_{W^p}(\gamma(s),\gamma(t))=|s-t|d_{W^p}(\mu,\nu)$. Is it true that $(Z,d)$ is also a geodesic space?
My Attempt: This should be true since $Z\ni z \mapsto \delta_{z} \in W_p$, here $\delta_z$ is the Dirac measure, is an isometric embedding, but I cannot fill out the details of the proof.
We know that a complete metric space is geodesic if and only if any two points $x,y$ have a midpoint $m$, i.e., $d(x,m)=d(y,m)=\frac{1}{2}d(x,y)$.
So, for any $z_0,z_1\in Z$, consider Dirac measures $\delta_{z_0},\delta_{z_1}$ and take their midpoint $\mu_m$(a probability measure). IF we could prove that $\mu_M=\delta_{z_m}$ for some $z_m\in Z$, then we have from the definition of a midpoint that
$d_{W^p}(\delta_{z_0},\delta_{z_m}) = d_{W^p}(\delta_{z_1},\delta_{z_m}) = \frac{1}{2}d_{W^p}(\delta_{z_0},\delta_{z_1})$
which is equivalent to $d(z_0,z_m)=d(z_1,z_m)=d(z_0,z_1)$ (since $z\to \delta_{z}$ is an isometry), proving that $z_m$ is a midpoint between $z_0,z_1$, so we have that $Z$ is a geodesic space.
However, I could not prove that $\mu_m$ is Dirac.
The implication usually goes the other way (see Corollary 7.2.2 of Villani's big yellow book: https://link.springer.com/book/10.1007/978-3-540-71050-9). For what you want, I am not sure it is true. Consider the two point space, $Z= \{-1,1\}$ equipped with the discrete metric (which is Polish as desired). I think its not possible to find a geodesic between $-1$ and $1$ but the curve $(1-t)\delta_{-1} + t\delta_1, t\in [0,1]$ is a geodesic between the corresponding Dirac measures in $W_1$. To see this note that the (optimal?) plan between $(1-t)\delta_{-1} + t\delta_1$ and $(1-s)\delta_{-1} + s\delta_1$ for $s<t$ is given by $(1-t)\delta_{(-1,-1)} + (t-s)\delta_{(-1,1)} + s \delta_{(1,1)}$ at least for $W_1$ and this satisfies your condition for being a geodesic. Note that since, linear combinations of Diracs are the only possible measures, we can construct geodesics between all measures in a similarly straightforward manner.
Presumably, a similar construction is possible for $p>1$, but I am not sure. I have not checked this very carefully so if I were you, I would accept what have written with an appropriate amount of suspicion...