  Let $\F$ be a finite family of axis-parallel boxes in $\R^d$ such that $\F$
contains no $k+1$ pairwise disjoint boxes. We prove that if $\F$ contains a
subfamily $\M$ of $k$ pairwise disjoint boxes with the property that for every
$F\in \F$ and $M\in \M$ with $F \cap M \neq \emptyset$, either $F$ contains a
corner of $M$ or $M$ contains $2^{d-1}$ corners of $F$, then $\F$ can be
pierced by $O(k)$ points. One consequence of this result is that if $d=2$ and
the ratio between any of the side lengths of any box is bounded by a constant,
then $\F$ can be pierced by $O(k)$ points. We further show that if for each two
intersecting boxes in $\F$ a corner of one is contained in the other, then $\F$
can be pierced by at most $O(k\log\log(k))$ points, and in the special case
where $\F$ contains only cubes this bound improves to $O(k)$.
