It seems to me that you've gone round in a circle. What's wrong with this simple argument?

Suppose that the fluid exerts a force $F_n$ on a small area $A$ of the container wall, in a direction normal to that area. If that area moves outwards by a small distance $\Delta x$ normal to $A$ then the work done by the fluid on $A$ will be $$w=F_n \Delta x=\frac{F_n}{A} \times A \Delta x = p \Delta V.$$ We are not in any way assuming that $w$ is a differential of some function of state, so there is no suspicion that $w$ is an exact differential.

It seems to me that you've gone round in a circle. What's wrong with this simple argument?

Suppose that the fluid exerts a force $F_n$ on a small area $A$ of the container wall, in a direction normal to that area. If that area moves outwards by a small distance $\Delta x$ normal to $A$ then the work done by the fluid on $A$ will be $$\delta w=F_n \Delta x=\frac{F_n}{A} \times A \Delta x = p \Delta V.$$ We are not in any way assuming that $\delta w$ is a differential of some function of state, so there is no suspicion that $\delta w$ is an exact differential.