Let $A,B$ be two sets such that $2A \cong 2B$ (here $2A := A \coprod A$). Then $A \cong B$. This can be proven without the axiom of choice, which means that one can explicitly construct a bijection $A \to B$ out of a bijection $2A \to 2B$. This is non-trivial and interesting, see the wonderful paper by Conway, Doyle, also for generalizations. The construction is infinitary, and therefore the following question comes into my mind.
Question. Is the assertion also true in ZF - {Axiom of Infinity}?