Do $\mathcal{U}$-small partitioned assemblies densely generate realizability toposes?

I don't know if this is what the authors of the paper had in mind, but here's one way to do it. Also, I'm not sure if there's a constructive way to do this, so I'll give an answer assuming the axiom of choice.

We take $\mathbb{C}_A$ to be the category of partitioned assemblies whose underlying set has cardinality less than or equal to that of $A$. Suppose we are given a partitioned assembly $(X, \phi)$, i.e. $X$ is a set and $\phi : X \to A$. We need to check that it is a colimit of $(Z, \psi)$ over the diagram of maps $(Z, \psi) \to (X, \phi)$. That is, we need to check that for all cocones over the diagram, say with apex $(Y, \rho)$, we can find a unique map $f : (X, \phi) \to (Y, \rho)$ making the necessary diagram commute. Certainly if we take the forgetful functor to sets, then the diagram is a colimit. Hence we do have a unique function $f : X \to Y$, and only need to show that it is tracked. We suppose that it is not tracked and get a contradiction. For each $a \in A$, since $a$ does not track $f$, there exists $x_a \in X$ such that $a \cdot \phi(x_a) \neq \rho(f(x_a))$. We then take $Z \subseteq X$ to be $\{x_a \;|\; a \in A\}$. This has cardinality less than or equal to $A$ by construction and we can make it a partitioned assembly by restricting $\phi$, with a canonical inclusion map $(Z, \phi) \to (X, \phi)$. However, the corresponding component of the cocone $(Z, \phi) \to (Y, \rho)$ cannot be tracked by any $a \in A$, giving a contradiction.

We can show that $\mathbb{C}_A$ is separating by similar argument: Suppose we have $f, g : U \rightrightarrows V$ in $\mathrm{RT}(A)$ such that $f \neq g$. Then $U$ is covered by a partitioned assembly, say $p : (X, \phi) \twoheadrightarrow U$. We have $f \circ p \neq g \circ p$. Hence there is no realizer for $f \circ p = g \circ p$, i.e. for all $a \in A$, $a$ does not realize $f \circ p = g \circ p$. Hence for each $a$ there is $x_a \in X$ such that $a \cdot \phi(x_a)$ does not realize $f(p(x_a)) = g(p(x_a))$. We again take $Z$ to be $\{x_a \;|\; a \in A\}$ and restrict $\phi$ to get a partitioned assembly in $\mathbb{C}_A$ that separates $f$ and $g$.

I think similar arguments should even show $\mathbb{C}_A$ is dense in $\mathrm{RT}(A)$, but it seems trickier.