Several rectangles cover the unit square. Can I find a disjoint set of them whose area is at least $1/4$? I am interested in the following question:

Let a finite sequence of rectangles in $\mathbb{R}^2$ be given such that

*

*The edges of the rectangles are parallel to the coordinate axes, and


*The rectangles cover the unit square, $[0,1]^2$.
Is it possible to find, among these rectangles, a collection of mutually disjoint rectangles whose combined area is at least $1/4$?

As of yet, I'm not sure if a solution exists. My friend and I have spent a while thinking about this and have gotten nowhere.
Any help is greatly appreciated.
 A: The question linked by @AlonAmit in the comments answers exactly this question, and shows that the answer (at least with the constant $1/4$) is no.  For a concrete demonstration, start with a $6\times 6$ square broken into thirty-sixths:
$$
\begin{matrix}
0&1&2&3&4&5\\
6&7&8&9&a&b\\
c&d&e&f&g&h\\
i&j&k&l&m&n\\
o&p&q&r&s&t\\
u&v&w&x&y&z
\end{matrix}
$$
Now cover each corner $2\times2$ by four individual $(1+\varepsilon)\times (1+\varepsilon)$ rectangles, such that the four rectangles in the upper left ($0,1,6,7$) are mutually overlapping, as are those in each of the other corners.  And cover the remaining shape in the center (a cross) by eight individual $(3+\varepsilon)\times(1+\varepsilon)$ rectangles ($28e$, $39f$, $cde$, $ijk$, $kqw$, $lrx$, $fgh$, and $lmn$), such that all eight include the center of the square.  Any disjoint set of these rectangles includes at most four of the $(1+\varepsilon)\times(1+\varepsilon)$ rectangles and at most one of the $(3+\varepsilon)\times(1+\varepsilon)$ rectangles, and so has total area just over $7/36\approx 19.4\%$ of the full square.
