Do light and heavy water form an azeotrope?
This is a good question, but no, there is no azeotrope in any isotopic mixture, as an azeotrope requires non-ideal mixing of two components to form a  minimum or maximum critical point on the plot of total vapor pressure as a function of composition as shown in Figure 1 below. Isotopic Mixtures have a near-zero enthalpy of mixing, making isotopic mixtures the closest thing to an ideal liquid mixture. Because of this ideality of mixing the boiling of protium-deuterium oxide mixtures looks much like the phase diagram in Figure 2 which follows Raoult's law, though this is actually quite exaggerated compared to semi-heavy water mixtures.  Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of $\ce{D2O}$ or $\ce{H2O}$ and be unobservable due to the "doped" species existing as semi-heavy water ($\ce{DHO}$).  This is further evidenced by the fact that other physical properties of isotopic mixtures follow a linear curve when measured as a function of concentration.

 
Figure 1. Generic Plot of vapor pressure as a function of composition

 
Figure 2. Generic Boiling phase diagram for ideal solutions

To answer your question about distillation only enrichment, it is possible though impractical.  Commercial enrichment of $\ce{D2O}$ first uses hydrogen sulfide in the Gridler-Sulfide process, once enriched to around 30%, distillation is used to finish the process to the desired isotopic purity.