Is the following argument correct to show that if every set $S$ can be well-ordered, then for any disjoint indexed family of non-empty sets $(X_{\alpha})_{\alpha \in A}$ there exists a map $f: \{ X_{\alpha} : \alpha \in A \} \to \cup_{\alpha \in A} X_{\alpha}$ such that $f(X_{\alpha}) \in X_{\alpha}$?
To show the claim, let $f(X_{\alpha}) = \min X_{\alpha}$. Then $f(X_{\alpha}) \in X_{\alpha}$, that is $f$ is the choice function, which exists because $X_{\alpha}$ has a well-ordering $\leq_{\alpha}$ in which $X_{\alpha}$ has a minimum.
Is this correct?
EDIT: It is assumed that the aforementioned $S$ is also non-empty.
You haven't said what is $S$. Namely you need to first fix a well ordering for $\bigcup X_\alpha$, only then the notion of a minimum makes sense.
Other than that, it's fine.