Proof. Let $\mathcal{F}$ be a disjoint family of nonempty sets. Let $\displaystyle f \colon \mathcal{F} \to \bigcup \mathcal{F}$ be a choice function. Let $A,B \in \mathcal{F}$ with $A \neq B$ Since $\mathcal{F}$ is a disjoint family of sets, $\displaystyle A \cap B = \emptyset$ Since $f$ is a choice function, $f(A) \in A$ and $f(B) \in B$ Thus, $f(A) \notin B$ Hence, $f(A) \neq f(B)$ It follows that $f$ is injective.

Let $\displaystyle C=\left\{f(B) \in \bigcup \mathcal{F} : B \in \mathcal{F} \right\}$ Then $C$ is a set.

Let $A \in \mathcal{F}$ Since $f$ is injective, $\displaystyle A \cap C=\{f(A)\}$