Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member? I consider definability to mean one of either cases:


*

*Definability without parameters (in the language of set theory), or

*Definability from ordinals and a real (in the same language).
So my question is:
Is there a model $M$ of ZFC (or at least of ZF) such that every definable family of sets (not necessarily of reals) contains at least one definable member (in the sense 1. or 2. respectively) but such that $M$ contains nonetheless many non-definable members (i.e. $M$ is not pointwise definable)?
 A: Yes,
fix a definable relation $\le_L$ that well-orders all of $L$.
If $V=L$ then every definable nonempty set $A$ has a definable member $a$, namely:
$a :=$ the $\le_L$-least element of $A$.
A: The answer in case 1. is also yes.  In fact, a stronger assertion is true: there exist models of set theory in which every set is definable without parameters.  Such models are called pointwise definable, and (as a first observation) are necessarily countable.  A collection of results surrounding pointwise definable models of ZFC and GBC (in which every set and every class are definable without parameters) are presented in "Pointwise Definable Models of Set Theory," joint work by Joel Hamkins, David Linesky and me. Here's a link to Joel's blog post on the paper, which gives an overview & link to the paper itself:  http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/
A: The following theorem seems to express how the various
definability witness properties are connected with each other and
with $V=\text{HOD}$.
Theorem. The following are equivalent in any model $M$ of ZF:


*

*$M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.

*$M$ has a definable well-ordering of the universe.

*Every definable nonempty set in $M$ has a definable element.

*Every definable nonempty set in $M$ has an ordinal-definable
element.

*Every $\Pi_2$-definable nonempty set in $M$ has an
ordinal-definable element.

*Every ordinal-definable nonempty set in $M$ has an
ordinal-definable element.
Proof. ($1\to 2$) The usual HOD order is a definable well-ordering
of the universe.
($2\to 3$) Select the least element with respect to the definable
order, as in Bjorn's answer.
($3\to 4$) Immediate.
($4\to 5$) Immediate.
($4\to 1$) If $M$ thinks there is a non-OD set, then the set $A$
of all non-OD sets in $M$ of minimal rank is a definable nonempty
set in $M$ with no ordinal-definable elements.
($5\to 1$) The stronger implication has now undergone a few
improvements, so let me discuss it. I had proposed considering as
above the set $A$ of all minimal-rank non-OD sets, which is
definable and nonempty in any model of $V\neq\text{HOD}$, but has
no ordinal-definable elements. I had guessed that $\Sigma_5$ would be
sufficient to define $A$. In the comments, François refined this, arguing that this set was actually 
$\Sigma_3$-definable and indeed $\Delta_3$-definable. Using his
idea, I was able to push this down to show that $A$ is
$\Sigma_2\wedge\Pi_2$ definable, by the properties: $A$ is not
empty; all elements of $A$ have the same rank; every element of
$A$ is not in OD; every set of rank less than an element of $A$ is
in OD; every set not in $A$, but of the same rank as an element of
$A$, is in OD. Each of these properties is either $\Sigma_2$ or
$\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable.
Specifically, the first two requirements are $\Sigma_2$, being
witnessed in a rank-initial segment of the universe; the third is
$\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are
true just in case there is a large $V_\theta$ which believes them
to be true. I also noted that $A$ is not provably $\Sigma_2$-definable.
Meanwhile, over at my question Can $V\neq\text{HOD}$ if every
$\Sigma_2$-definable set has an ordinal-definable element?, Emil made a suggestion
leading to the observation that if $V\neq\text{HOD}$, then there
is a $\Pi_2$-definable set with no ordinal-definable elements. The
set is simply $U=A\times V_\theta$, where $A$ is as above and $\theta$ is
least such that $V_\theta$ thinks $A$ is the set of minimal-rank
non-OD sets. So I refer the reader to theorem 2 in that answer,
which provides the content of the implication ($5\to 1$).
($1\to 6$) Immediate, since under statement $1$, every set in $M$
is ordinal-definable in $M$.
($6\to 4$) Immediate. QED
Conclusion. Thus, case (1) of the question occurs in exactly
the models of $V=\text{HOD}$ that are not pointwise definable.
There are such models, if ZFC is consistent, since one may take
any uncountable model of $\text{ZFC}+V=\text{HOD}$.
Meanwhile, case (2) of the question — ignoring the issue of
real parameters — does not occur at all, since if a set has
sets that are not ordinal-definable, then it will have a definable
set with no ordinal-definable members, namely, the set of all
non-OD sets of minimal rank, as in the implication of statement 4
to statement 1.
Update. I edited to the improved statement 5, which we've now
got down to the case of mere $\Pi_2$-definability, using the
answer to my question Can $V\neq\text{HOD}$ if every
$\Sigma_2$-definable set has an ordinal-definable element?.
Update. This answer and those of the related questions have known grown into the following paper:

F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? (arχiv:1706.07285)
Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.
Read more at the blog post.

