Yes. You're correct. I realised this some time after writing my comment. So it actually does work as a definition, if you phrase it properly. I guess it's jus

Yes. You're correct. I realised this some time after writing my comment. So it actually does work as a definition, if you phrase it properly. I guess it's just equivalent to the observation that ω has the property that every initial segment (i.e. every subset of the form {0, 1, ..., n}) is finite. So you could say: a set S is countable if it admits an ordering that makes it 'covered by all finite initial segments'. ...but this boils down to saying the set admits an ordering that makes it isomorphic to some n, or ω... which just means it's countable in the usual sense.