If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.
One way for an irrational number $\alpha$ to be in this new set is because I threw away something like (?, $\alpha$) and ($\alpha$, ?). Each interval I removed can account for at most two numbers in this manner, and since I threw away countably many intervals, this only explains countably many irrationals that remain.
But since my [0,1] set without the cover has measure 1 - $\epsilon$, there must be more than countably many points.
So there must be another way for irrationals to remain than how I described above; but I can't think of how.