[proofplan]
We prove uniqueness by contradiction-free comparison of two arbitrary minimizers. The uniform convexity hypothesis on $V$ gives strict displacement convexity of the free energy along every $W_2$-geodesic, via [citetheorem:9568]. Applying this inequality to a geodesic joining two minimizers shows that any positive Wasserstein distance between them would produce an intermediate measure with energy strictly below the minimum. Hence the endpoint distance is zero, and since $W_2$ is a metric on $\mathcal P_2(\mathbb R^n)$, the two minimizers coincide.
[/proofplan]
[step:Choose two minimizers with the same finite energy value]
Define the minimal value $m\in\mathbb R$ by
\begin{align*}
m:=\inf_{\mu\in\mathcal P_2(\mathbb R^n)}\mathcal F[\mu].
\end{align*}
By hypothesis, $m<+\infty$ and there exists at least one $\mu_*\in\mathcal P_2(\mathbb R^n)$ with $\mathcal F[\mu_*]=m$.
Let $\mu_0,\mu_1\in\mathcal P_2(\mathbb R^n)$ be arbitrary minimizers of $\mathcal F$. Then
\begin{align*}
\mathcal F[\mu_0]=m
\end{align*}
and
\begin{align*}
\mathcal F[\mu_1]=m.
\end{align*}
In particular, both endpoint energies are finite.
[/step]
[step:Apply strict displacement convexity along a Wasserstein geodesic]
Since $(\mathcal P_2(\mathbb R^n),W_2)$ is a geodesic metric space, choose a constant-speed $W_2$-geodesic $\mu_\bullet:[0,1]\to\mathcal P_2(\mathbb R^n)$, $t\mapsto\mu_t$, from $\mu_0$ to $\mu_1$. Since $\mu_0$ and $\mu_1$ have finite free energy and $D^2V(x)=J(\nabla V)_x\ge\lambda I_n$ for every $x\in\mathbb R^n$, [citetheorem:9568] applies to this geodesic and gives, for every $t\in[0,1]$,
\begin{align*}
\mathcal F[\mu_t]\le (1-t)\mathcal F[\mu_0]+t\mathcal F[\mu_1]-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
Using $\mathcal F[\mu_0]=\mathcal F[\mu_1]=m$, this becomes
\begin{align*}
\mathcal F[\mu_t]\le m-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
[guided]
Define
\begin{align*}
m:=\inf_{\mu\in\mathcal P_2(\mathbb R^n)}\mathcal F[\mu].
\end{align*}
Since $\mu_0$ and $\mu_1$ are minimizers, we have $\mathcal F[\mu_0]=m$ and $\mathcal F[\mu_1]=m$. We now use the strict convexity information supplied by [citetheorem:9568]. The map $\mu_\bullet:[0,1]\to\mathcal P_2(\mathbb R^n)$, $t\mapsto\mu_t$, is chosen to be a constant-speed $W_2$-geodesic with endpoints $\mu_0$ and $\mu_1$, where $W_2$ is the quadratic Wasserstein distance declared in the theorem statement. The endpoint hypotheses needed for [citetheorem:9568] are satisfied because
\begin{align*}
\mathcal F[\mu_0]=m<+\infty
\end{align*}
and
\begin{align*}
\mathcal F[\mu_1]=m<+\infty.
\end{align*}
The space $(\mathcal P_2(\mathbb R^n),W_2)$ is geodesic, so there exists a constant-speed $W_2$-geodesic $\mu_\bullet:[0,1]\to\mathcal P_2(\mathbb R^n)$, $t\mapsto\mu_t$, from $\mu_0$ to $\mu_1$. The convexity hypothesis is exactly the stated lower bound
\begin{align*}
D^2V(x)=J(\nabla V)_x\ge\lambda I_n
\end{align*}
for every $x\in\mathbb R^n$.
Therefore, for every $t\in[0,1]$, the displacement convexity estimate gives
\begin{align*}
\mathcal F[\mu_t]\le (1-t)\mathcal F[\mu_0]+t\mathcal F[\mu_1]-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
Since both endpoints are minimizers, their energies are both equal to the same number $m$. Substituting these two equalities into the right-hand side yields
\begin{align*}
\mathcal F[\mu_t]\le (1-t)m+tm-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
The affine part simplifies to $m$, so
\begin{align*}
\mathcal F[\mu_t]\le m-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
This is the point where strictness enters: if the two minimizers were separated by positive $W_2$-distance, the last term would lower the energy below $m$ for every $t\in(0,1)$.
[/guided]
[/step]
[step:Use minimality to force the Wasserstein distance to vanish]
Fix $t\in(0,1)$. Since $\mu_t\in\mathcal P_2(\mathbb R^n)$ and $m$ is the infimum of $\mathcal F$ over $\mathcal P_2(\mathbb R^n)$, we have
\begin{align*}
m\le \mathcal F[\mu_t].
\end{align*}
Combining this with the previous estimate gives
\begin{align*}
m\le \mathcal F[\mu_t]\le m-\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
Subtracting $m$ from the leftmost and rightmost terms gives
\begin{align*}
0\le -\frac{\lambda}{2}t(1-t)W_2(\mu_0,\mu_1)^2.
\end{align*}
Since $\lambda>0$ and $t(1-t)>0$, this is possible only if
\begin{align*}
W_2(\mu_0,\mu_1)^2=0.
\end{align*}
Thus
\begin{align*}
W_2(\mu_0,\mu_1)=0.
\end{align*}
[/step]
[step:Conclude equality of the minimizers and finite free energy]
The function $W_2$ is a metric on $\mathcal P_2(\mathbb R^n)$, so the identity-of-indiscernibles property gives
\begin{align*}
\mu_0=\mu_1.
\end{align*}
Because $\mu_0$ and $\mu_1$ were arbitrary minimizers, $\mathcal F$ has at most one minimizer. The hypothesis that $\mathcal F$ attains its finite minimum gives at least one minimizer $\mu_*$, so this minimizer is unique. Finally,
\begin{align*}
\mathcal F[\mu_*]=m<+\infty,
\end{align*}
so the unique minimizer has finite free energy.
[/step]
