I have enough musings to post them as an answer, rather than fill
up comment space.

First : Use a simple recursive construction to get
an upper bound on the supremum. This places x_1 at 1,
x_2n at x_n - 1/2n, and x_(2n+1) at x_n + 1/(2n+1). This
gives an upper bound of sum{i positive integer} 1/(2^i - 1)
which is some number less than 169/105. Of course, you need
to prove this construction works.

Second: viewed as a tree with node n branching to children
x_(2n) and x_(2n+1), note that you can prune and graft the
tree, reshaping it as needed. Specifically, start by
exchanging branches at nodes 11 and 7. (This works because
1/2 + 1/5 + 2/7 < 2* 1/2 = 1.) You may find that repruning
smaller branches leads more quickly to a near optimal bound.
Even with the one graft made, the upper bound is reduced to
less than 1147/759.

Third: start determining optimal placements for the first
n terms for small n, which meet the conditions and stay
below the bounds established above. A computer
simulation should quickly run through placements for n
up to 12 which stay below the lower bound. For example,
by hand one sees that x_1 < x_2 < x_n for n < 80 already
leads to non optimal placements, so that combined with
some analysis should prove that x_2 < x_1 in an optimal
placement.

This approach should lead you quickly to the first four
decimal digits of the supremum.

**UPDATE 07.11 **: I have what I think are two tools
to tackle the problem. The first tool is the bounded width
branch: Given n, form the branch suggested above starting
with x_n "representing" 1/n, placing x_2n at x_n - 1/2n and
x_2n+1 at x_n + 1/(2n+1), and continuing recursively. The
actual tool is the lemma that this branch meets the criteria
for extending the sequence and does so using up at most 2/n
space, and actually at most 1/n + 1/(2n+1) + 1/(4n+3) + ... .

Formally the lemma should read: Let for j in S be
the subsequence described above, where n in S is given
and for k in S one has both 2k and 2k+1 in S, and no other
integers or objects are in S otherwise. This subsequence
can be part of a sequence that satisfies the spacing
criterion given in the problem, and max(x_i - x_j) for
i,j coming from S is less than 2/n.

The second tool is that, given any starting sequence,
there is a way to extend it using bounded width branches
to get a solution. Formally: Let for m <= M be
a finite subsequence which satisfies the spacing criterion
given. Then there are M+1 bounded width branches that can
be grafted on to the sequence, given a complete sequence
that also satisfies the spacing requirements.

Proof sketch: start with x_M, and place x_2M and x_2M+1
adjacent to it. Then go backwards up to x_M+1, placing
bounded width branches in the space next to the smallest
undecorated leaf. The spacing requirements guarantee that
the branches will fit without needing to move any of the
first M x_i . Also, show that the branches aren't close
enough to each other to conflict with the spacing requirement.

So for any suitable sequence of length M, one can extend
it to a complete suitable sequence at a cost of at most
2/(M+1). Now with this estimate, one can go through the
first few finite sequences and weed out those that are
provably nonoptimal.

END UPDATE 07.11

Gerhard "Ask Me About System Design" Paseman, 2010.07.08