If the air is cooled down, then the rate at which molecules leave the liquid slows down. The molecules entering the liquid do not slow down at the same rate, causing the liquid to grow toward it's initial state.

Note that I specifically said it is a vacuum. Instead of a glass of water, picture the water as little drops. The atmosphere can act to warm or cool these drops, and vice-versa.

In the more nitty-gritty aspect of this, the equation that describes the vapor pressure as a function of temperature is called the Clausius-Clapeyeron equation/relation. The American Meteorological Society has one approximate solution, but I prefer this equation: $$e_{sat}(T)=611 Pa \exp[\frac{L_v}{R_v}(273.15^{-1}-T^{-1})]$$, where $L_v$ is the latent heat of vaporization, $R_v$ is the specific gas constant for water vapor, and $T$ is the absolute temperature in Kelvin. Combined with the ideal gas law for water vapor (assuming saturation) $$e_{sat}(T)V=m_vR_vT$$, and given the volume ($V$) we can write an expression for the mass of water vapor $m_v$. The equation comes out to $$m_v=611 Pa \exp[\frac{L_v}{R_v}(273.15^{-1}-T^{-1})]V R_v^{-1}T^{-1}$$

To answer your final question, the molecules are approximated as being infintessimally small, per the ideal gas law. To be more specific, one molecule of water is about 7.08$\times$ 10$^{-19}$ cubic feet (after some math), so the added volume is considered negligible. In short, the molecules are treated as point masses.

Raising the temperature of the solid/liquid and of the vapor means fewer vapor molecules will be moving slow enough to be captured while more solid/liquid will be moving fast enough to escape. This increases the pressure at which equilibrium occurs. Lowering the temperature similarly decreases the pressure at which equilibrium occurs.