Here's one example that requires neither measure theory nor an enumeration of the rational numbers. The set of positive irrationals is in bijection with the set of infinite sequences of positive integers through their continued fraction representation (which is unique for irrationals). Take the subset for which all of the terms of the continued fraction representation are $1$ or $2$. The complement in the set of positive reals is open, since for every irrational that has some continued fraction coefficient${}>2$ one easily finds a neighborhood where the coefficients up to and including this one are unchanged, and in a sufficiently small neighborhood of every rational number $\alpha$, some continued fraction coefficient (beyond the finite representation of $\alpha$ itself) is necessarily very big.
I may add that if the set ${\Bbb N_{>0}}^\Bbb N$ of infinite sequences of positive integers is equipped with the product topology for the discrete topology of $\Bbb N_{>0}$ (so that subsets of sequences in which a finite initial subsequence is fixed form a basis of open sets), then the continued fraction representation is a homeomorphism of the positive irrationals (with their topology restricted from $\Bbb R$) to ${\Bbb N_{>0}}^\Bbb N$. Therefore this is just a concrete version of the answer by Brian M. Scott (but with positive irrationals only, and with a more elementary argument to show the set is closed).