[proofplan]
We identify the stated curve with the standard displacement interpolation associated to the optimal plan $\pi=(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0$. The only real work is checking that pushing $\mu_0$ forward by $x \mapsto (1-t)x+tT(x)$ is the same as pushing $\pi$ forward by $(x,y)\mapsto (1-t)x+ty$. Once this identification is made, the general displacement interpolation theorem for an optimal quadratic-cost plan gives the constant-speed $W_2$ geodesic property, and the endpoint identities follow from $F_0=\operatorname{id}_{\mathbb{R}^n}$ and $F_1=T$.
[/proofplan]
[step:Define the interpolation maps and verify the curve lies in $\mathcal{P}_2(\mathbb{R}^n)$]
For each $t \in [0,1]$, define the Borel map
\begin{align*}
F_t:\mathbb{R}^n \to \mathbb{R}^n,\qquad F_t(x)=(1-t)x+tT(x).
\end{align*}
Since $T$ is Borel and affine combinations preserve Borel measurability, $F_t$ is Borel. Hence $\mu_t=(F_t)_\#\mu_0$ is a Borel probability measure on $\mathbb{R}^n$.
We verify that $\mu_t$ has finite second moment. Since $\pi=(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0$ is a coupling of $\mu_0$ and $\mu_1$, its second marginal is $\mu_1$, and therefore
\begin{align*}
T_\#\mu_0=\mu_1.
\end{align*}
Using the elementary estimate $|a+b|^2 \leq 2|a|^2+2|b|^2$ in $\mathbb{R}^n$, with $a=(1-t)x$ and $b=tT(x)$, we obtain
\begin{align*}
\int_{\mathbb{R}^n} |z|^2\,d\mu_t(z)=\int_{\mathbb{R}^n}|F_t(x)|^2\,d\mu_0(x).
\end{align*}
Thus
\begin{align*}
\int_{\mathbb{R}^n} |z|^2\,d\mu_t(z)\leq 2(1-t)^2\int_{\mathbb{R}^n}|x|^2\,d\mu_0(x)+2t^2\int_{\mathbb{R}^n}|T(x)|^2\,d\mu_0(x).
\end{align*}
Since $T_\#\mu_0=\mu_1$, the change-of-variables formula for pushforward measures gives
\begin{align*}
\int_{\mathbb{R}^n}|T(x)|^2\,d\mu_0(x)=\int_{\mathbb{R}^n}|y|^2\,d\mu_1(y).
\end{align*}
Both terms on the right are finite because $\mu_0,\mu_1\in\mathcal{P}_2(\mathbb{R}^n)$. Hence $\mu_t\in\mathcal{P}_2(\mathbb{R}^n)$ for every $t\in[0,1]$.
[/step]
[step:Identify the map-induced curve with displacement interpolation from the optimal plan]
Define, for each $t\in[0,1]$, the Borel map
\begin{align*}
G_t:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n,\qquad G_t(x,y)=(1-t)x+ty.
\end{align*}
We claim that
\begin{align*}
(G_t)_\#\pi=(F_t)_\#\mu_0.
\end{align*}
Indeed, since $\pi=(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0$, the functoriality of pushforwards under composition gives
\begin{align*}
(G_t)_\#\pi=(G_t)_\#(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0=(G_t\circ(\operatorname{id}_{\mathbb{R}^n},T))_\#\mu_0.
\end{align*}
For every $x\in\mathbb{R}^n$,
\begin{align*}
(G_t\circ(\operatorname{id}_{\mathbb{R}^n},T))(x)=G_t(x,T(x))=(1-t)x+tT(x)=F_t(x).
\end{align*}
Therefore
\begin{align*}
(G_t)_\#\pi=(F_t)_\#\mu_0=\mu_t.
\end{align*}
[guided]
The goal is to show that the formula in the theorem is not a new construction: it is exactly the usual displacement interpolation formula, specialized to the case where the optimal plan is concentrated on the graph of a transport map.
For each $t\in[0,1]$, define
\begin{align*}
G_t:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n,\qquad G_t(x,y)=(1-t)x+ty.
\end{align*}
The map $G_t$ is continuous, hence Borel measurable. This map takes a pair consisting of a starting point $x$ and an ending point $y$ and returns the point lying a fraction $t$ of the way from $x$ to $y$. The displacement interpolation associated to a transport plan $\pi$ is precisely $(G_t)_\#\pi$.
Here the plan is induced by the map $T$, so
\begin{align*}
\pi=(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0.
\end{align*}
Define the graph map $A:\mathbb{R}^n\to\mathbb{R}^n\times\mathbb{R}^n$ by $A(x)=(x,T(x))$. Since $T$ is Borel and $\operatorname{id}_{\mathbb{R}^n}$ is continuous, $A$ is Borel measurable. Pushing $\mu_0$ forward by this map records the starting point and its transported endpoint.
Now use the composition rule for pushforward measures: if $\nu$ is a measure and $A,B$ are measurable maps with compatible domains, then $B_\#(A_\#\nu)=(B\circ A)_\#\nu$. The required measurability has just been checked: $A$ is Borel and $G_t$ is Borel. Applying this with $\nu=\mu_0$, the graph map $A$, and $B=G_t$, we get
\begin{align*}
(G_t)_\#\pi=(G_t)_\#A_\#\mu_0=(G_t\circ A)_\#\mu_0.
\end{align*}
The composed map is computed pointwise:
\begin{align*}
(G_t\circ A)(x)=G_t(x,T(x))=(1-t)x+tT(x)=F_t(x).
\end{align*}
Therefore the two pushforwards agree:
\begin{align*}
(G_t)_\#\pi=(F_t)_\#\mu_0=\mu_t.
\end{align*}
This is the key identification: the map formula in the statement is exactly the displacement interpolation formula applied to the graph plan.
[/guided]
[/step]
[step:Apply displacement interpolation for an optimal quadratic-cost plan]
By hypothesis, $\pi$ is an optimal transport plan between $\mu_0$ and $\mu_1$ for the quadratic cost $c(x,y)=|x-y|^2$. Let $\Pi(\mu_0,\mu_1)$ denote the set of Borel probability measures on $\mathbb{R}^n\times\mathbb{R}^n$ whose first marginal is $\mu_0$ and whose second marginal is $\mu_1$. The displacement interpolation theorem for an optimal quadratic-cost plan states that if $\gamma\in\Pi(\mu_0,\mu_1)$ is optimal and
\begin{align*}
\nu_t=(G_t)_\#\gamma,\qquad G_t(x,y)=(1-t)x+ty,
\end{align*}
then $(\nu_t)_{t\in[0,1]}$ is a constant-speed $W_2$ geodesic from $\mu_0$ to $\mu_1$.
The theorem applies with $\gamma=\pi$, because $\pi$ is a coupling of $\mu_0$ and $\mu_1$ and is optimal by assumption. From the previous step, $\nu_t=(G_t)_\#\pi=\mu_t$ for every $t\in[0,1]$. Hence $(\mu_t)_{t\in[0,1]}$ is a constant-speed $W_2$ geodesic. In particular, for all $s,t\in[0,1]$,
\begin{align*}
W_2(\mu_s,\mu_t)=|s-t|W_2(\mu_0,\mu_1).
\end{align*}
[/step]
[step:Check the endpoint measures]
At $t=0$, the interpolation map is
\begin{align*}
F_0(x)=x
\end{align*}
for every $x\in\mathbb{R}^n$, so $F_0=\operatorname{id}_{\mathbb{R}^n}$ and
\begin{align*}
\mu_{0,\mathrm{curve}}=(F_0)_\#\mu_0=(\operatorname{id}_{\mathbb{R}^n})_\#\mu_0=\mu_0.
\end{align*}
At $t=1$, the interpolation map is
\begin{align*}
F_1(x)=T(x)
\end{align*}
for every $x\in\mathbb{R}^n$, so
\begin{align*}
\mu_{1,\mathrm{curve}}=(F_1)_\#\mu_0=T_\#\mu_0.
\end{align*}
Since $\pi=(\operatorname{id}_{\mathbb{R}^n},T)_\#\mu_0$ is a coupling of $\mu_0$ and $\mu_1$, its second marginal is $\mu_1$, which is exactly the statement that $T_\#\mu_0=\mu_1$. Therefore the curve starts at $\mu_0$ and ends at $\mu_1$. Together with the constant-speed identity proved above, this completes the proof.
[/step]
