Let $(X,d)$ be a totally bounded metric space and $\mathcal F(X)$ its set of nonempty finite subsets equipped with the following metric:
$$ d_W(A, B) = \!\!\! \sup_{\substack{f:X\to\mathbb{R}\\\text{1-Lipschitz}}} \left| \frac{1}{|A|}\sum_{a\in A}f(a) - \frac{1}{|B|}\sum_{b\in B}f(b) \right|. $$
My question: Can we show constructively that $(\mathcal F(X),d_W)$ is totally bounded directly from the total boundedness of $X$?
By "constructively" I mean something like this: For any $\delta>0$ we know that $X=\bigcup_{i=1}^n B^d_\delta(x_i)$ for some $\{x_1,\ldots,x_n\}\subseteq X$. Can we use these points to construct (for each $\epsilon>0$) a finite collection of sets $\{A_1,\ldots,A_m\}\subseteq \mathcal F(X)$ such that $\mathcal F(X)=\bigcup_{i=1}^m B_\epsilon^{d_W}(A_i)$ by directly using the formula for $d_W$?
If we replace $d_W$ with the Hausdorff distance $d_H$ then the power set works. I suspect the power set suffices for $d_W$ as well but have been unable to show it.
Example of "non-constructive" proof:
Identifying $A\in\mathcal F(X)$ with the empirical measure $\mu_A=\tfrac{1}{|A|}\sum_{a\in A}\delta_a$ reveals that $d_W$ is just the 1-Wasserstein metric $W$ restricted to the empirical measures via the Kantorovich-Rubinstein formula:
$$ W(\mu_A, \nu_A) = \!\!\! \sup_{\substack{f:X\to\mathbb{R}\\\text{1-Lipschitz}}} \left| \int f\, d\mu_A - \int f\,d\nu_A \right| =d_W(A,B). $$
Since the completion $\overline X$ is compact, general theory tells us the space of probability measures $\mathcal P(\overline X)$ is compact. By slight abuse of notation, we then see that $\mathcal F(X)\subseteq \mathcal P(\overline X)$ is totally bounded because it is a subset of a compact metric space.
