Volume change in positive deviation from Raoult's Law
I'm a little late to the party here, but wanted to chime in.

The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured.  Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form.  Imagine an open beaker of ethanol with a small disk floating on top so that no molecules can escape upwards, but it can move up and down with the change in liquid level.  This would also keep the pressure of the liquid constant (although it doesn't really matter since the liquid is essentially incompressible) .

If you were able to inject a certain amount of acetone into this beaker without opening it to the atmosphere for very long, the resulting volume of liquid in the beaker would be greater than the sum of its parts. Hopefully it makes sense that if there is no space for liquid to form, the positive volume change of mixing can show itself.

As far as the idea that more vapor would form to decrease the volume of liquid, realize that evaporation is a time-dependent process, so likely the mixed liquid would reach an effective equilibrium state where it's volume change could be observed before the extra liquid has time to evaporate.  It is true that it would evaporate faster if the effective vapor pressure (technically "fugacity" or "chemical potential") were higher, as the driving force for mass transfer would be greater, but this is also very dependent on the way the system is specified.  Technically if you left the container open in a large room forever, it would all evaporate in either case.