Actually, all that is required for g(m,n), m,n countable, is for g(m,n) to be a mapping, any mapping, even countably multi-valued, ie, g(m,n) = +/-sqrt(mn), mn positive and countable.
A countable union of countable sets is countable. You should really carefully follow the proof in Rudin for this.
On the other hand the introduction of g in this thread has been rendered redundant by the very simple fact that (m,n,p,q) is countable answers the original op that rational points in the plane are countable. See my post # above.
This result for g developed in the course of the thread. I suspect the management disapproves of developing a topic in the course of a thread. That's unfortunate. Interesting things come out of it. This post also belongs at the end of the companion thread "map (m,n) to N" from which I have been unsubscribed, after slipEternal had his say.
So feel free to respond to this post without fear of successful contradiction because I probably won't be able to respond. Personally, I think it's an interesting development.