In an attempt to write a proof by contradiction, I end up with a space $X$ with the following properties:


(0) $X$ is nonempty,  

(1) $X$ is Hausdorff,  

(2) $X$ has no isolated points,  

(3) every subspace of $X$ is constructible (finite union of locally closed subsets).



> 
> Is this indeed a contradiction? 
> 
> 
> 


It would suffice to know that any $X$ with properties (1) and (2) has a dense subset with dense complement: such a set cannot be constructible unless $X=\emptyset$. [Edit: there are counterexamples to this, see the comment by Yves Cornulier]



Constructible sets in Hausdorff spaces
No, it is not a contradiction.


A space $X$ is called *submaximal* if every subset of $X$ is locally closed (and hence constructible). It is easy to see that spaces with only finitely many non-isolated points are submaximal. Finding a submaximal Hausdorff space with no isolated points seems harder but there are plenty of those too.


It is a nice exercise to show that any maximal space is in fact submaximal. A Hausdorff space $(X,\tau)$ is called *maximal* if it has no isolated points but any finer topology $\tau' \supset \tau$ has isolated points. With a standard use of Zorn´s lemma you can show that any topology without isolated points is contained in a maximal topology. This gives you a way of "constructing" a lot of submaximal Hausdorff spaces with no isolated points. Some of them are even Tychonoff spaces (see E. van Douwen, "*Applications of maximal topologies*", Topology Appl., 1993), although these are a lot harder to get.


