In the 3rd edition of the Principles of Mathematical Analysis by Rudin, he defined a countable set as one which has a bijection between the set of Natural numbers and itself ie. it is infinite. He also defined a separable metric space as a set which possessed a countably dense set. Does this mean that all separable metric spaces have infinite dense subsets?
The reason for this question is the following:
A metric space in which every infinite set has a limit point is separable
In the solution provided in the link above and other solutions I've seen, a countably dense set was constructed but their definition of countable includes finite sets as well. No effort was shown to determine if the set needed to be infinite.
So for this particular question shown i Take countable as referring to finite sets as well?
Certainly every infinite separable metric space has a countably infinite dense subset. Let $ X $ be your space, $ D $ a countable dense subset (which exists by separability) and $ A $ a countably infinite subset (which exists as $ X $ is infinite). Then $ D \cup A $ is a countably infinite dense subset of $ X $.
As a general rule, when a question refers to a countable set, it should never require finiteness, as countable always at least includes the possibility of countably infinite. Thus, the distinction is often irrelevant, especially in the case of separable metric spaces as you can always perform the above maneuver to restrict to the countably infinite case.