If two sets $A,B$ are taken from collection $\mathcal C$ and $A\neq B$ implies in that situation that $A\cap B=\emptyset$ then the (non-indexed) collection $\mathcal C$ can be labeled as 'pairwise disjoint'. Looked at as an indexed collection more is needed: $i\neq j$ must imply that $A_i\cap A_j=\emptyset$
Underlying is the fact that you can have two distinct indices $i\neq j$ with $A_i=A_j$.
Example:
Take $A_i$ for $i=1,2,3$ and $A_1=A_2\neq\emptyset$ and $A_1\cap A_3=\emptyset$.
Then collection $\{A_1,A_2,A_3\}=\{A_1,A_3\}$ is pairwise disjoint, but indexed collection $\{A_i\}_{i\in\{1,2,3\}}$ is not.
I have never seen "pairwise disjoint family indexed by $I$" used to mean Definition 1; it always means Definition 2 in my experience. Note though that you can also talk about a pairwise disjoint family where "family" just means "set" rather than "indexed set". In that case, Definition 1 is what it means for the set (not the indexed set) $\{A_i\}_{i\in I}$ to be pairwise disjoint. Incidentally, I strongly recommend not using the same notation for a set and an indexed set; I would use $(A_i)_{i\in I}$ for the indexed set (which is really a function with domain $I$) and $\{A_i\}_{i\in I}$ for the set (which is the image of that function). On a related note, if you are using "family" to mean indexed set rather than just set, your statement of the axiom of choice should be
If $(A_i)_{i\in I}$ is a family of pairwise disjoint, non-empty sets then there is $C$ such that for all $i\in I$, $|A_i\cap C|=1$.
The version you stated cannot be used with either of your definitions since neither definition defines what it means for a set $A$ (rather than an indexed set $(A_i)_{i\in I}$) to be a pairwise disjoint family. (Of course, you can make such a definition, and get a third version of the axiom of choice. It is trivial to see that this third version is equivalent to the axiom of choice using Definition 2, since given any non-indexed family $A$ you can make it an indexed family by the identity map $A\to A$ (that is, you take $I=A$ and $A_i=i$).)
In any case, the axiom of choice for Definition 1 is equivalent for the axiom of choice for Definition 2. The forward direction is trivial; for the reverse direction, assume the axiom of choice for Definition 2 and let $(A_i)_{i\in I}$ be a family of pairwise disjoint sets by Definition 1. Let $J=\{A_i:i\in I\}$ and consider the indexed set $(B_j)_{j\in J}$ defined by $B_j=j$.
I claim $(B_j)_{j\in J}$ is a pairwise disjoint family according to Definition 2. Indeed, suppose $j,j'\in J$ and $j\neq j'$. By definition of $J$, there exist $i,i'\in I$ such that $j=A_i$ and $j'=A_{i'}$. Then $A_i\neq A_{i'}$ and $i\neq i'$ (since $i\mapsto A_i$ is a function), so since $(A_i)_{i\in I}$ satisfies Definition 1, $A_i\cap A_{i'}=\emptyset$. But $A_i=j=B_j$ and $A_{i'}=j'=B_{j'}$, so $B_j\cap B_j=\emptyset$.
Thus by the axiom of choice for Definition 2, there exists a set $C$ such that $|B_j\cap C|=1$ for each $j\in J$. Now for any $i\in I$, let $j=A_i$, so $A_i=B_j$. We thus see that $|A_i\cap C|=|B_j\cap C|=1$. So, this set $C$ also has the required property for the family $(A_i)_{i\in I}$.
Best Answer
There is no need to use contradiction. Suppose that $B_0,B_1$ in $\mathscr{B}$ with $B_0\ne B_1$. Then $B_0$ and $B_1$ are distinct members of $\mathscr{A}$, so $B_0\cap B_1=\varnothing$ (since $\mathscr{A}$ is a pairwise disjoint family). Thus, every pair of distinct members of $\mathscr{B}$ are disjoint, and by definition $\mathscr{B}$ is a pairwise disjoint family.
The argument that you’ve given does not make sense. $\mathscr{B}$ is a family of sets, and in order to show that this family is pairwise disjoint, you must show that if $B_0,B_1\in\mathscr{B}$, then either $B_0=B_1$, or $B_0\cap B_1=\varnothing$. There is no point to looking at $\mathscr{A}\cap\mathscr{B}$: we know that $\mathscr{A}\cap\mathscr{B}=\mathscr{B}$ simply because $\mathscr{B}\subseteq\mathscr{A}$. This fact has in itself no bearing on whether $\mathscr{B}$ is pairwise disjoint.