I don't think there is a constructive proof like for the Sogenfrey line, at least if the rational sequences are not given beforehand. If we merely consider two disjoint subsets of the irrationals we can arrange for sequences with even denominators to converge to the first set, and for sequences with odd denominators to converge to the other and then basic neighborhoods suffice to separate the sets.

This is an excellent example! Is there an explicit construction of two closed, disjoint sets that cannot be separated in this space?

This does necessarily work. You can fix a bijection from 2^Q to R such that {q} is mapped to q, and the usual ordering of R is total, but Q is not well ordered...

I know. My question is if the following are equivalent: 1. Axiom of choice 2. For every mapping $f : X -> Y$ that is not continuous at $x_0$ there is a net converging to $x_0$ that does not converge to $f(x_0)$ It is not hard to prove that 1 implies 2. As you have shown, to prove 2 you don't even have to assume 1 if the space is $T_1$ (the question however is not settled for arbitrary topological spaces)

This is a good answer. I am still interested in the other direction. The proof that Tychonoff's theorem, or that every vector space has a basis imply the axiom of choice - I find very fascinating and I wonder if similar techniques can be applied in this case.

It still proves the claim that a finite collection of functions can't satisfy the criteria so I think you should edit your post to emphasize that

I don't understand the last paragraph. What if each $A_n$ happens to be set of all those reals larger than $n$?

@Joel - yes. I have not even considered that option :) @oktan - You are totally right, but that's easily fixed by extending the list to an infinite set consisting of constant functions

An easy construction of an almost disjoint family of size continuum is given by taking, for each real number with, say, decimal expansions 0.123411... the set {1, 12, 123, 1234, ... }.