TITLE: Is there a countable, regular space with no isolated points which is not homeomorphic to the rationals?
QUESTION [9 upvotes]: It's known that every countable, metrizeable space with no isolated points is homeomorphic to the rationals with the standard topology.
Suppose you wanted to reformulate the above without referencing metrizability directly. Since a countable space is clearly separable, you can use Urysohn's characterization of separable metric spaces to replace metrizeable with 2nd countable, regular (I take regular to mean regular and Hausdorff). You could even get away with 1st countable, regular since, for a countable space, 1st countable implies 2nd countable. 
My question is, what happens if we try to relax metrizability to regularity? Is the space automatically 1st countable (and hence homeomorphic to the rationals)? To summarize:

Is there a countable space $X$ which is regular Hausdorff with no isolated points, but not 1st countable?

The Arens-Fort space (example 26 from Counterexamples in Topology) shows there is a countable, regular Hausdorff (in fact $T_5$) space which is not 1st countable. Unfortunately, the space has lots of isolated points.

REPLY [4 votes]: Another naturally occurring example is the following: by the Hewitt-Marczewksi theorem, the set $I^I$, where $I$ is the unit interval, in the product topology, is a separable space, so has a countable dense subset $D$. As the product $I^I$ is completely regular, so is $D$, and so it's normal (being Lindelöf) and thus hereditarily normal, even. But at no point can $D$ have a local base of size $< \mathfrak{c}$ (very non-isolated), as is shown in Engelking's book, IIRC (no copy at hand). 
As an aside: I have sometimes wondered: are all such countable dense subsets homeomorphic?