You can actually show that 2 is an upper bound. To do this, we will show that:

[Existence of a valid subset] $$$\iff$$$ [There exists a pair of rows $$$(i, j)$$$ where $$$(i & j) = 0$$$]. Here, $$$i$$$ and $$$j$$$ are the numbers encoded in the set bits of the rows.

Backward direction: This is easy to see.

Forward direction: Suppose we have a valid subset with $$$k$$$ rows, where no row consists entirely of zeroes (otherwise we can simply take that row and be done):

Lemma: There exists a row in this subset with at least 3 zeroes.

Proof: We know that each column in the subset has at least $$$\lceil\frac{k}{2}\rceil$$$ zeroes. We know that there are 5 columns. So, there are at least $$$5\lceil\frac{k}{2}\rceil$$$ zeroes in total. Also, $$$5\lceil\frac{k}{2}\rceil > 2k$$$, which proves the statement.

Now, suppose we choose a row from our subset with at least 3 zeroes. There are 2 cases:

1. The chosen row has 4 zeroes. In this case, the column without the zero must have another row with a zero. We can choose that other row and have a valid pair.
2. The chosen row has 3 zeroes. There are now 2 columns without zeroes. So, in the remaining $$$(k - 1)$$$ rows, both of these columns must have $$$\lceil\frac{k}{2}\rceil$$$ zeroes. Since $$$2\lceil\frac{k}{2}\rceil > (k - 1)$$$, there must be another row where both of these columns are zero.

What does this mean? The original problem can be solved simply by finding any pair of rows with a bitwise-AND of zero. This is not too hard for $$$n = 5$$$, and for slightly bigger $$$n$$$ we can use SOS DP.