[proofplan]
Let $\pi$ be the unique optimal coupling. The standard displacement interpolation theorem gives existence: pushing $\pi$ forward by the affine maps $e_t(x,y)=(1-t)x+ty$ produces a constant-speed $W_2$ geodesic. For uniqueness in the stated class, take any other straight-line interpolation induced by an optimally paired coupling $\gamma$; the stated endpoint cost equality says exactly that $\gamma$ realizes the endpoint transport cost, hence is optimal. Since the optimal coupling is unique, $\gamma=\pi$, and applying the same maps $e_t$ gives the same intermediate measures. In the map-induced case, the identity $e_t\circ(\operatorname{id},T)=(1-t)\operatorname{id}+tT$ gives the displayed formula.
[/proofplan]
[step:Construct the displacement interpolation from the unique optimal plan]
Let $\pi \in \Pi(\mu_0,\mu_1)$ denote the unique optimal plan for the quadratic cost. For each $t \in [0,1]$, define
\begin{align*}
e_t: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n, \quad e_t(x,y) = (1-t)x + ty.
\end{align*}
Define the curve
\begin{align*}
\mu: [0,1] \to \mathcal{P}_2(\mathbb{R}^n), \quad t \mapsto \mu_t := (e_t)_\#\pi.
\end{align*}
Since $\mu_0,\mu_1 \in \mathcal{P}_2(\mathbb{R}^n)$ and $\pi \in \Pi(\mu_0,\mu_1)$ is an optimal quadratic-cost plan, the cost of $\pi$ is finite and equals $W_2(\mu_0,\mu_1)^2$. The maps $e_t$ move each pair $(x,y)$ along the Euclidean line segment from $x$ to $y$ at constant speed. Therefore the standard displacement interpolation theorem applies to the optimal coupling $\pi$ and implies that $\mu$ is a constant-speed $W_2$ geodesic from $\mu_0$ to $\mu_1$ (citing a result not yet in the wiki: Displacement interpolation along an optimal plan is a $W_2$ geodesic). Indeed, $e_0(x,y)=x$ and $e_1(x,y)=y$, so
\begin{align*}
\mu_0 = (e_0)_\#\pi, \quad \mu_1 = (e_1)_\#\pi.
\end{align*}
[/step]
[step:Associate an endpoint coupling to any competing straight-line geodesic]
Let $\gamma \in \Pi(\mu_0,\mu_1)$ be a coupling satisfying
\begin{align*}
\int_{\mathbb{R}^n \times \mathbb{R}^n} |x-y|^2 \, d\gamma(x,y) = W_2(\mu_0,\mu_1)^2.
\end{align*}
Define the curve
\begin{align*}
\nu: [0,1] \to \mathcal{P}_2(\mathbb{R}^n), \quad t \mapsto \nu_t := (e_t)_\#\gamma.
\end{align*}
We will record that $\gamma$ is optimal.
[/step]
[step:Use the endpoint cost equality to prove the endpoint coupling is optimal]
By the hypothesis on the competing optimally paired coupling $\gamma$,
\begin{align*}
\int_{\mathbb{R}^n \times \mathbb{R}^n} |x-y|^2 \, d\gamma(x,y) = W_2(\mu_0,\mu_1)^2.
\end{align*}
Since $\gamma \in \Pi(\mu_0,\mu_1)$ and its quadratic cost equals the infimum defining $W_2(\mu_0,\mu_1)^2$, $\gamma$ is an optimal plan.
[guided]
We now explain why the endpoint cost hypothesis forces the coupling $\gamma$ to be optimal. The phrase "optimally paired particles" means precisely that $\gamma$ is a coupling of $\mu_0$ and $\mu_1$ whose quadratic endpoint cost realizes the squared Wasserstein distance:
\begin{align*}
\int_{\mathbb{R}^n \times \mathbb{R}^n} |x-y|^2 \, d\gamma(x,y) = W_2(\mu_0,\mu_1)^2.
\end{align*}
By definition, $W_2(\mu_0,\mu_1)^2$ is the infimum of the same quadratic cost over all couplings in $\Pi(\mu_0,\mu_1)$. Since $\gamma$ belongs to $\Pi(\mu_0,\mu_1)$ and attains this infimum, $\gamma$ is an optimal plan. This is the exact point where optimal pairing is used; without this endpoint cost equality, a straight-line particle interpolation may have geodesic-looking marginals only under additional hypotheses, and the coupling itself need not have been shown optimal.
[/guided]
[/step]
[step:Invoke uniqueness of the optimal plan to identify all intermediate measures]
The optimal plan between $\mu_0$ and $\mu_1$ is unique by hypothesis, and the previous step proves that $\gamma$ is optimal. Hence
\begin{align*}
\gamma = \pi.
\end{align*}
For every $t \in [0,1]$, applying the same Borel map $e_t$ to both sides gives
\begin{align*}
\nu_t = (e_t)_\#\gamma = (e_t)_\#\pi = \mu_t.
\end{align*}
Thus every constant-speed $W_2$ geodesic obtained by Euclidean constant-speed transport of paired particles agrees with the displacement interpolation induced by the unique optimal plan.
[/step]
[step:Rewrite the interpolation when the optimal plan is induced by a map]
Assume now that there exists a Borel map $T: \mathbb{R}^n \to \mathbb{R}^n$ such that
\begin{align*}
\pi = (\operatorname{id},T)_\#\mu_0.
\end{align*}
For each $t \in [0,1]$, define
\begin{align*}
S_t: \mathbb{R}^n \to \mathbb{R}^n, \quad S_t(x) = (1-t)x + tT(x).
\end{align*}
Then, for every $x \in \mathbb{R}^n$,
\begin{align*}
(e_t \circ (\operatorname{id},T))(x) = e_t(x,T(x)) = (1-t)x + tT(x) = S_t(x).
\end{align*}
Using functoriality of pushforwards under composition,
\begin{align*}
\mu_t
=
(e_t)_\#\pi
=
(e_t)_\#(\operatorname{id},T)_\#\mu_0
=
(e_t \circ (\operatorname{id},T))_\#\mu_0
=
(S_t)_\#\mu_0.
\end{align*}
Therefore
\begin{align*}
\mu_t = ((1-t)\operatorname{id}+tT)_\#\mu_0, \quad t \in [0,1].
\end{align*}
This is the asserted formula and completes the proof.
[/step]
