[proofplan]
The proof is a finite telescoping argument. At each JKO step, we test the minimizing property of $\rho_{k+1}^\tau$ against the previous iterate $\rho_k^\tau$, which is an admissible competitor because the whole sequence lies in $\mathcal A$. This gives a one-step energy dissipation inequality, and summing those inequalities from $k=0$ to $N-1$ cancels all intermediate energy terms.
[/proofplan]
[step:Test each minimizing step against the previous iterate]
Fix $k\ge 0$. Define the penalized functional $\Psi_k:\mathcal A\to(-\infty,\infty]$ by
\begin{align*}
\Psi_k[\rho]=\mathcal F[\rho]+\frac{1}{2\tau}W_2^2(\rho,\rho_k^\tau).
\end{align*}
By hypothesis, $\rho_{k+1}^\tau$ minimizes $\Psi_k$ over $\mathcal A$. Since $(\rho_j^\tau)_{j=0}^{\infty}\subset\mathcal A$, the element $\rho_k^\tau$ is an admissible competitor. Therefore
\begin{align*}
\mathcal F[\rho_{k+1}^\tau]+\frac{1}{2\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_k^\tau]+\frac{1}{2\tau}W_2^2(\rho_k^\tau,\rho_k^\tau).
\end{align*}
Since $W_2$ is a metric on $\mathcal P_2(\mathbb R^n)$, one has $W_2(\rho_k^\tau,\rho_k^\tau)=0$. Hence
\begin{align*}
\mathcal F[\rho_{k+1}^\tau]+\frac{1}{2\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_k^\tau].
\end{align*}
[guided]
Fix $k\ge 0$. The minimizing property is useful only after choosing a competitor in the same admissible class. Define the penalized functional $\Psi_k:\mathcal A\to(-\infty,\infty]$ by
\begin{align*}
\Psi_k[\rho]=\mathcal F[\rho]+\frac{1}{2\tau}W_2^2(\rho,\rho_k^\tau).
\end{align*}
The hypothesis says precisely that $\rho_{k+1}^\tau$ is a minimizer of this functional over $\mathcal A$. Thus, for every competitor $\rho\in\mathcal A$, one has $\Psi_k[\rho_{k+1}^\tau]\le \Psi_k[\rho]$.
We choose the competitor $\rho=\rho_k^\tau$. This is allowed because the theorem assumes $(\rho_j^\tau)_{j=0}^{\infty}\subset\mathcal A$, so in particular $\rho_k^\tau\in\mathcal A$. Substituting this competitor into the minimizing inequality gives
\begin{align*}
\mathcal F[\rho_{k+1}^\tau]+\frac{1}{2\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_k^\tau]+\frac{1}{2\tau}W_2^2(\rho_k^\tau,\rho_k^\tau).
\end{align*}
The last term on the right vanishes because $W_2$ is a metric on $\mathcal P_2(\mathbb R^n)$, and every metric satisfies $d(x,x)=0$. Hence $W_2(\rho_k^\tau,\rho_k^\tau)=0$, and the one-step dissipation estimate becomes
\begin{align*}
\mathcal F[\rho_{k+1}^\tau]+\frac{1}{2\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_k^\tau].
\end{align*}
This is the basic inequality that will telescope after summation.
[/guided]
[/step]
[step:Sum the one-step inequalities and telescope the energies]
Let $N\in\mathbb N$. Summing the one-step inequality over $k=0,\dots,N-1$ gives
\begin{align*}
\sum_{k=0}^{N-1}\mathcal F[\rho_{k+1}^\tau]+\frac{1}{2\tau}\sum_{k=0}^{N-1}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \sum_{k=0}^{N-1}\mathcal F[\rho_k^\tau].
\end{align*}
Because all the displayed energy values are finite by hypothesis, subtracting the finite common sum $\sum_{k=1}^{N-1}\mathcal F[\rho_k^\tau]$ from both sides is legitimate. The remaining terms are
\begin{align*}
\mathcal F[\rho_N^\tau]+\frac{1}{2\tau}\sum_{k=0}^{N-1}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_0^\tau].
\end{align*}
Finally, $\rho_0^\tau=\rho_0$ by hypothesis, so $\mathcal F[\rho_0^\tau]=\mathcal F[\rho_0]$. Therefore
\begin{align*}
\mathcal F[\rho_N^\tau]+\frac{1}{2\tau}\sum_{k=0}^{N-1}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)\le \mathcal F[\rho_0].
\end{align*}
This is the desired finite-step discrete energy dissipation inequality.
[/step]
