[proofplan]
The proof uses only the variational definition of the discrete minimizing movement. At each time step, compare the minimizer $x_{k+1}^\tau$ with the admissible competitor $x_k^\tau$ to obtain a one-step energy inequality. Summing those one-step inequalities over $k=m,\dots,\ell-1$ makes the intermediate energy terms telescope, leaving exactly the stated discrete dissipation estimate.
[/proofplan]
[step:Compare each minimizer with the previous time step]
Fix $k\in\mathbb N_0$. Define the penalized functional $\Phi_k:X\to(-\infty,+\infty]$ by
\begin{align*}
\Phi_k(y)=\mathcal E(y)+\frac{1}{2\tau}d^2(y,x_k^\tau).
\end{align*}
By the discrete minimizing movement hypothesis, $x_{k+1}^\tau$ minimizes $\Phi_k$ over $X$. Since $x_k^\tau\in X$ and $\mathcal E(x_k^\tau)<+\infty$, it is an admissible competitor. Hence
\begin{align*}
\mathcal E(x_{k+1}^\tau)+\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)\le \mathcal E(x_k^\tau)+\frac{1}{2\tau}d^2(x_k^\tau,x_k^\tau).
\end{align*}
Because $d(x_k^\tau,x_k^\tau)=0$, this becomes
\begin{align*}
\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_{k+1}^\tau)\le \mathcal E(x_k^\tau).
\end{align*}
[guided]
Fix a time index $k\in\mathbb N_0$. The definition of a discrete minimizing movement says that $x_{k+1}^\tau$ is chosen by minimizing the previous energy plus a quadratic penalty for moving away from $x_k^\tau$. To use this definition, introduce the map $\Phi_k:X\to(-\infty,+\infty]$ by
\begin{align*}
\Phi_k(y)=\mathcal E(y)+\frac{1}{2\tau}d^2(y,x_k^\tau).
\end{align*}
The hypothesis says precisely that $\Phi_k(x_{k+1}^\tau)\le \Phi_k(y)$ for every $y\in X$.
We choose the competitor $y=x_k^\tau$. This point is allowed because $x_k^\tau\in X$ and, by hypothesis, $\mathcal E(x_k^\tau)<+\infty$. Therefore
\begin{align*}
\Phi_k(x_{k+1}^\tau)\le \Phi_k(x_k^\tau).
\end{align*}
Expanding the definition of $\Phi_k$ gives
\begin{align*}
\mathcal E(x_{k+1}^\tau)+\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)\le \mathcal E(x_k^\tau)+\frac{1}{2\tau}d^2(x_k^\tau,x_k^\tau).
\end{align*}
The metric axiom $d(x_k^\tau,x_k^\tau)=0$ removes the penalty term on the right-hand side. Thus the one-step dissipation inequality is
\begin{align*}
\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_{k+1}^\tau)\le \mathcal E(x_k^\tau).
\end{align*}
This is the entire source of the estimate: each minimizing step pays for its movement by decreasing the energy.
[/guided]
[/step]
[step:Sum the one-step inequalities over the desired time interval]
Let $m,\ell\in\mathbb N_0$ satisfy $0\le m<\ell$. Summing the one-step inequality from $k=m$ to $k=\ell-1$ gives
\begin{align*}
\sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\sum_{k=m}^{\ell-1}\mathcal E(x_{k+1}^\tau)\le \sum_{k=m}^{\ell-1}\mathcal E(x_k^\tau).
\end{align*}
Reindexing the second sum on the left gives
\begin{align*}
\sum_{k=m}^{\ell-1}\mathcal E(x_{k+1}^\tau)=\sum_{j=m+1}^{\ell}\mathcal E(x_j^\tau).
\end{align*}
Thus
\begin{align*}
\sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\sum_{j=m+1}^{\ell}\mathcal E(x_j^\tau)\le \sum_{k=m}^{\ell-1}\mathcal E(x_k^\tau).
\end{align*}
Subtracting the common finite sum $\sum_{j=m+1}^{\ell-1}\mathcal E(x_j^\tau)$ from both sides yields
\begin{align*}
\sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_\ell^\tau)\le \mathcal E(x_m^\tau).
\end{align*}
Since
\begin{align*}
\sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)=\frac{1}{2}\sum_{k=m}^{\ell-1}\frac{d^2(x_{k+1}^\tau,x_k^\tau)}{\tau},
\end{align*}
the desired estimate follows.
[/step]
