Is every proportionately dissociated set a finite union of dissociated sets? (Erdős #774)

Call $A\subset\mathbb{N}$ dissociated if $\sum_{n\in X}n\neq\sum_{m\in Y}m$ for all finite $X,Y\subset A$ with $X\neq Y$ (equivalently, all finite subset sums of $A$ are distinct). Call an infinite set $A\subset\mathbb{N}$ proportionately dissociated if there is a constant $c>0$ such that every finite $B\subset A$ contains a dissociated subset of size $\ge c\lvert B\rvert$. Alon and Erdős asked: is every proportionately dissociated set the union of finitely many dissociated sets?

Acceptance. FULLY RESOLVES: a complete proof (Lean/Coq preferred, else full written proof). NO: an explicitly described $A\subset\mathbb{N}$ with a proof that it is proportionately dissociated (exhibit the constant $c$ and the argument) yet is not the union of any finite number of dissociated sets. YES: a proof that every proportionately dissociated set decomposes into finitely many dissociated sets, with an explicit bound on the number of pieces in terms of the constant $c$. ADVANCES: a proof for a restricted class (e.g. lacunary-type sets or sets of bounded additive energy); a proof that the number of dissociated pieces, if finite, cannot be bounded solely in terms of $c$; or a verified finite construction realising a record 'gap' (a finite set whose every dissociated subset is small yet which requires many dissociated pieces), with the search code and certificate. Deliver the proof, the counterexample set with its proportionate-dissociation certificate, or the finite construction plus code.