Clausius–Clapeyron relation for states of water and vapor I've seen many questions about this topic on this site, but I still have some issues trying to understand this aspect. Here's what I believe I understood
looking at the phase diagram for a pure substance (water):


If you put water in a container with a piston pushing $\pu{1 atm}$ directly on the water with no extra space for inert gases, it will stay 100% liquid as at room temperature and $\pu{1 atm}$ liquid is the stable phase.
If the water is put at room temperature in a closed box where the volume is more than the volume of the water (creating a vacuum), the molecules will evaporate in a number so that the water vapor will be at the equilibrium pressure corresponding to room temperature, which can be calculated by the phase diagram.


I can't understand why the air (and its pressure) doesn't work like a piston, making sure no water is in gas state at $\pu{1 atm}$.

In particular, there was this exercise where a piston with $p = p_\mathrm{tot}$ was pushing down on a mixture of vapor $\ce{A}$ and inert gas above pure liquid $\ce{Ar}$, so that:

\begin{align}
p_\mathrm{tot} &amp;= p_\mathrm{inert} + p_\mathrm{a}\\
\text{Equilibrium:}\qquad \ce{A(l, $p_\mathrm{tot}$) &amp;= Inert(g, $p_\mathrm{inert}$) + Vapor(g, $p_\mathrm{a}$)}
\end{align}

Why doesn't the vapor $\ce{A}$ 'feel' the $p_\mathrm{tot}$ and get condensed into liquid $\ce{A}$?

Was it because $p_\mathrm{tot}$ is less than the vapour pressure of pure $\ce{A}$ at that temperature? If it was $p_\mathrm{tot} = p_\mathrm{vap}$ ($T = T_\mathrm{exercise}$), then the equilibium would be

$$\ce{A(pure, l, $p_\mathrm{tot}, T$) = Inert(g, $T, p_\mathrm{tot}$)?}$$

This doesn't look like an everyday case to me though where water is in gas form even if the pressure is $\pu{1 atm}$.
The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal).
Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result is as expected evaporation or sublimation of the substance until the pressure of the vapour has increased to an equilibrium value (the saturation vapor pressure). We assume there is plenty of substance so that some remains in the condensed phase while the rest fills the box as vapour. What drives evaporation/sublimation? The fact that the chemical potential of the substance is (initially) lower in the vapour. Once equilibrium is achieved the chemical potential is the same in both phases.
Now assume an alternate scenario: in addition to the substance (solid or liquid) there is now some inert gas in the box. As suspected this gas exerts a pressure on the liquid and thereby affects its chemical potential (free energy). But the free energy of the substance in the gas phase is unaffected by the inert gas (assuming the gases behave ideally). And here the somewhat counterintuitive result: the substance again evaporates/sublimes until reaching the saturation vapour pressure, but precisely because the inert gas exerts a pressure on the condensed phase and increases its chemical potential in this phase, but does not alter the chemical potential of the vapor, it increases the saturation vapor pressure. The increased vapour pressure raises the chemical potential of the vapour until it matches the heightened potential of the condensed phase.
And the short answer to the question is that the inability of vapor A to 'feel' the total pressure including the inert gas is part of the definition of an ideal or perfect gas, which fails if the pressure is raised sufficiently.