I am reading a book on set theory and I got stuck on a simple step of a proof involving an equivalence of the axiom of choice.
Axiom of Choice (1) For every set $a\neq \emptyset$ there exists a function $f:\mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}\to a$ such that $f\left (b\right )\in b$ for every $b\subseteq a$, $b\neq \emptyset$
Axiom of Choice (2) Let $\left \{a_i\right \}_{i\in I}$ be a non-empty family of pairwise disjoint non-empty sets (i.e. $I\neq \emptyset$ and $a_i\cap a_j=\emptyset$ if $i\neq j$). Then there exists a set $d$ such that $d\cap a_i$ is a unit set for every $i\in I$.
I could understand the proof $\left (2\right )\Rightarrow \left (1\right )$, but I cannot understand the converse.
The proof starts by supposing that $\left (1\right )$ is true. We let $\left \{a_i\right \}_{i\in I}$ be a non-empty family of pairwise disjoint non-empty sets and define $a=\bigcup_{i\in I}a_i\neq \emptyset$. Therefore there exists a function $f:\mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}\to a$ be a function such that $f\left (b\right )\in b$ for every $b\in \mathcal{P}\left (a\right )\setminus \left \{\emptyset\right \}$. If $d$ is the image of $f$, then (the book says) we have $d\cap a_i=\left \{f\left (a_i\right )\right \}$. My question is: Why? It is obvious that one of the inclusions holds, but what about $\subseteq$?
Recall that $a=\bigcup_{i\in I}a_i$ is trivially a superset of each of the $a_i$'s. Therefore each $a_i$ is an element of $\mathcal P(a)$.
By assumption, no $a_i$ is empty, so $a_i$ is in the domain of the choice function. Now take $\{f(a_i)\mid i\in I\}$ to be $d$, and by the fact each $a_i$ is disjoint we get that $d\cap a_i$ is a singleton for all $i\in I$.