Filter-closed vs. chain-closed Let A is a complete lattice.
I call a subset $S$ of A filter-closed when for every filter base $T$ in $S$ we have $\bigcap T\in S$. (A filter base is a nonempty, down directed set.)
I call a subset $S$ of A chain-closed when for every non-empty chain $T$ in $S$ we have $\bigcap T\in S$.
Conjecture $S$ is filter-closed if and only if $S$ is chain-closed.
 A: Indeed, your conjecture is correct. 
Theorem. If L is a complete lattice and S is a subset of L, then S is chain-closed iff S is filter-closed.
Proof. Clearly filter-closed implies chain-closed, since every chain is a filter base. Conversely, suppose that S is chain-closed, and that A is a filter base contained in S. Note that S is trivially filter-closed with respect to any finite filter base. So suppose by induction that S is filter-closed with respect to any filter base of size smaller than |A|. Enumerate A = { aα | α < |A| }. Let bβ be the meet of { aα | α < β }. This is the same as the meet of the filter sub-base of A generated by this set. This filter sub-base has size less than |A|, and hence by induction every bβ is in S. Also, the bβ are a descending chain in S, since as we take more aα, the meet gets smaller. Thus, by the chain-closure of S, the meet b of all the bβ is in S. This meet b is the same as the meet of A, and so we have proved that S is filter-closed. QED
This argument is very similar to the following characterization of (downward) complete lattices (which I had posted as my original answer). 
Theorem. The following are equivalent, for any
lattice L.


*

*L is complete, in the sense that every subset of L has a
greatest lower bound.

*L is filter complete, meaning that every filter base in
L has a greatest lower bound.

*L is chain complete, meaning that every filter base in L
has a greatest lower bound.
Proof. It is clear that completeness implies filter
completeness, since every filter base is a subset of L, and
filter completeness implies chain completeness, since every
chain is a filter base. For the remaining implication, suppose that L is chain
complete. We want to show that every subset A of L has a
greatest lower bound in L. We can prove this by transfinite
induction on the size of A. Clearly this is true for any
finite set, since L is a lattice. Fix any infinite set A.
Enumerate A as { aα | α < |A| }.
By the induction hypothesis, for each β < |A|, the
set { aα | α < β } has a
greatest lower bound bβ. Note that {
bβ | β < |A| } is a chain, because
as we include more elements into the sets, the greatest
lower bound becomes smaller. Thus, there is an element b in
L that is the greatest lower bound of the
bβ's. It is easy to see that this element b
is also a lower bound of A. QED
One can describe the method as finding a linearly ordered
cofinal sequence through the filter generated by the filter
base. This proof used AC when A was enumerated, and I
believe that this cannot be omitted.
One can modify the argument to show that for every infinite
cardinal κ, then a lattice is κ-complete (every
subset of size less than κ has a glb) iff every
filter base of size less than κ has a glb iff every
chain of size less than κ has a glb.
Note that if the lattice is bounded (meaning that it has
a least and greatest element), then having greatest lower
bounds for every set is the same as having least upper
bounds for every set, since the least upper bound of a set
A is the greatest lower bound of the set of upper bounds of
A. Thus, a complete lattice is often defined as saying
that every subset has a glb and lub.
There have been a few questions here at MO concerning
complete lattices. See this one and this one.
Questions about the degree of completeness of a partial
order often arise in connection with forcing arguments, and
when one is speaking of partial completeness and partial
orders (rather than lattices), and the situation is
somewhat more subtle. For example, a partial order P is
said to be κ-closed if every linearly ordered subset
of P of size less than κ has a lower bound. It is
κ-directed closed if every filter base in P of size
less than κ has a lower bound. With these concepts,
it is no longer true that a partial order is
κ-directed closed if and only if it is
κ-closed. One example arising in forcing would be the
forcing to add a slim κ-Kurepa tree, which is
κ-closed but not κ-directed closed. The
difference between these two concepts is related to
questions of large cardinal indestructibility, for Richard
Laver proved that every supercompact cardinal κ can
become indestructible by all κ-directed closed
forcing, but no such cardinal can ever be indestructible by
all κ-closed forcing, precisely because the slim
κ-Kurepa tree forcing destroys the measurability of
κ.
A: How is the following related to the question...
Let $P$ be a partially-ordered set.  Suppose every chain in $P$ has a least upper bound.  Then every subset of $P$ which is directed has a least upper bound.
I needed this once, long ago, didn't find it, so included a proof in the paper.  Many years later someone gave me a reference for it:  Mayer-Kalkschmidt & Steiner, Duke Math. J. 31 (1964) 287-289
A: I made a post to sci.math regarding this question.  You might check that post for some suggestions.  Others might check the thread for more history on that problem.
Gerhard "Ask Me About System Design" Paseman, 2010.02.18
A: I wrote more detailed proof based on the proof by Joel David Hamkins in this online article.
