Why isn't the half-disk topology separable? 
The half-disk topology is the topology on the upper half plane (with the $X$-axis) where base sets are open balls in the upper half plane and open half discs with the center point (on the $X$-axis). The exact definition is in the picture attached. (it's from the book counterexamples in topology). My problem is, that later in that page, the writer claims that this space is not
separable, and I can't see why. Take the set of all points (p.q) where $p$ and $q$ are rational numbers and where $q \geq 0$. Isn't this a countable dense set if the space described??
 A: This must be a mistake in the text.  Note that this is Example 78 in Counterexamples in Topology, and if you look at the charts in the Appendix it indicates that Example 78 is separable.  (And your reasoning appears to be impeccable to me!)
Probably what was meant is that the subspace $L$ of $X$ is not separable (it is uncountable and discrete!), and so the entire space cannot be second-countable (since every subspace of a second-countable space is second-countable, hence separable).

Addendum:  This appears to be an error in the first edition.  The above answer took information from the second edition of Counterexamples.  Upon closer inspection, the second edition's information about the Half-Disc Topology differs from the first's on the point in question:

(5.) The subspace $L$ is discrete and uncountable, so $( X , \tau^* )$ is not second countable. Neither is it Lindelöf, for the covering by basis neighbourhoods has no countable subcover.  But clearly $(X , \tau^* )$ is both separable and first countable.

Perhaps find a copy of the second edition?  (Cheap from Dover! Cheaper from Amazon!)
