If for any family of sets $\mathcal{F}$ we call $S$ a choice set of $\mathcal{F}$ iff there is a map $f:\mathcal{F}\to\cup_{S\in\mathcal{F}}S$ such that $\forall X\in\mathcal{F}(f(X)\in X)$ and $f[\mathcal{F}]=S$ we see the choice sets of any family $\mathcal{F}$ are just the images of choice functions of $\mathcal{F}$ - now with that said do these "choice sets" of $\mathcal{F}$ already go by another name?
Update: The choice sets of a family of sets $\mathcal{F}$ are always the feasible sets of an interval greedoid, more specifically when $M_{{\small \mathcal{F}}}=\{X\in\mathcal{F}:\forall S\in\mathcal{F}(S\not\subset X)\}$ is the family of all minimal sets in $\mathcal{F}$ where $T_{{\small \mathcal{F}}}$ is the family of all bases for the transversal matroid of $\mathcal{F}$, so if $\small [A,B]=\{S:A\subseteq S\subseteq B\}$, then $\small C=\{\phi[\mathcal{F}]:\phi\text{ is a choice function of }\mathcal{F}\}=\cup_{(X,Y)\in b(M_{{\small \mathcal{F}}})\times T_{{\small \mathcal{F}}}}[X,Y]$ is exactly the choice sets of $\small\mathcal{F}$ i.e. choice sets of $\mathcal{F}$ will form the interval greedoid whose family of minimal feasible sets is $b(M_{{\small \mathcal{F}}})$ (the blocker of $M_{{\small \mathcal{F}}}$) and whose maximal feasible sets are bases of the transversal matroid on $\mathcal{F}$. Don't see the point in writing this as an answer myself since Asaf devoted time so I'm accepting it.
Let me answer from the perspective of a set theorist. Since every set can be the image of a choice function (e.g. if $A$ is a set, it is the image of the only choice function from $\{\{a\}\mid a\in A\}$), so I am not aware of any particular terminology.
We do want to make a distinction when the family of sets is pairwise disjoint; in this case we can treat the family of sets as equivalence classes of an equivalence relation on some set. In this case the terms "selector", "transversal", or "system of representatives" are sometimes thrown around. And they make sense, too.