Piercing axis-parallel boxes

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 $O(k)$