Let $A$ be a (commutative) ring. A family $(B_i)_{i\in I}$ of $A$-algebras is said to be an fppf cover if it satisfies three properties: (1) each $B_i$ is flat as an $A$-module, (2) each $B_i$ is finitely presented as an $A$-algebra, and (3) the family is faithful, meaning that any $A$-module map $M\to N$ is an isomorphism if (and only if) each of the induced maps $B_i\otimes_A M \to B_i\otimes_A N$ is.

It is a theorem that for every fppf cover $(B_i)_{i\in I}$, there is a finite subfamily $J\subseteq I$ such that $(B_j)_{j\in J}$ is an fppf cover. In other words, affine schemes are quasi-compact in the fppf topology. The standard proof uses algebraic geometry: combine the fact that any flat finitely presented map of affine (say) schemes is open with the fact that affine schemes are quasi-compact in the Zariski topology.

My question is whether anyone knows a "direct" proof of this fact. Ideally it would not use the theorem that flat finitely presented morphisms are open or maybe not even any algebraic geometry at all. You might also hope that it works in some more general context, for instance for maps of (commutative?) monoids in any symmetric monoidal category.