Family of pairwise distinct elements Given a family $(x_i)_{i\in I}$ of (not necessarily pairwise distinct) elements, is it always possible to choose a subset $J\subset I$ such that
$$ \{x_i: i\in I\} = \{x_j: j \in J\} $$
and such that the elements $x_j$ with $j\in J$ are pairwise distinct?
Do we need the axiom of choice? Sorry if this should be obvious.
 A: I presume that you intend to require $J$ to "cover" the whole family in that for each $i\in I$, there exists $j\in J$ such that $x_i=x_j$.  The statement you ask for is then equivalent to the axiom of choice.
In one direction, assume the axiom of choice and let $(x_i)_{i\in I}$ be a family.  Let $S=\{x_i:i\in I\}$ and for each $s\in S$, let $I_s=\{i\in I:s\in S\}$.  Then $\{I_s:s\in S\}$ is a set of disjoint nonempty sets, so by the axiom of choice there exists a set $J$ that contains exactly one element from each $I_s$.  This exactly means that the terms of the family $(x_j)_{j\in J}$ are distinct but cover the entire original family $(x_i)_{i\in I}$.
Conversely, suppose it is always possible to pick such a subset $J\subseteq I$, and let $A$ be any set of disjoint nonempty sets; we wish to find a set that contains exactly one element of each element of $A$.  Let $I=\bigcup A$ and consider the family $(x_i)_{i\in I}$ defined by saying $x_i$ is the unique element of $A$ that contains $i$.  By hypothesis, there is then a subset $J\subseteq I$ such that the $x_j$ for $j\in J$ are distinct but cover all the values of $x_i$ for $i\in I$.  This means that $J$ contains exactly one element from each element of $A$, as desired.
