This follows a prior question:
What is the proof in Zermelo, without using choice nor foundation, of the cardinality of the Boolean union of two equal sized disjoint pure Dedekind infinite sets being equal to the cardinality of each one of them? Formally: $$ \forall A \forall B: A,B \text { are Ddk infinite} \land A \cap B = \emptyset \land |A|=|B| \\\implies |A \cup B| = |A|$$
Where: $$ A \text { is Ddk infinite } \iff \exists B \subsetneq A \exists f: A \to B, f \text { is injective}$$
Now, $$ A \text { is pure Ddk infinite } \iff \\\forall B \subseteq A (B \text { is Tarski infinite } \implies B \text { is Ddk infinite}) $$
You cannot prove this even in ZF. For instance, it is consistent with ZF+DC (in particular, with ZF+"every infinite set is Dedekind-infinite") that there exists an $\aleph_1$-amorphous set, which is a set such that every subset is either countable or cocountable. Such a set cannot be in bijection with a disjoint union of two copies of itself, since each copy in the disjoint union is uncountable and councountable.