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Here $A + B := \{(0, a) \ | \ a \in A\} \cup \{(1, b) \ | \ b \in B \}$.

Under the axiom of choice, we have that there exists an injection $f: B \to A$.

Because $g: A \to A + B: a \mapsto (0, a)$ is injective, it is enough to find an injection from $A + B$ to $A$ (Schröder-Cantor-Bernstein). I'm stuck on finding this injection.

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Pick an enumeration of a countable subset $S$ of $A$, say $x_1,x_2,\dots$. Enumerate $B$ as $y_1,y_2,\dots$. Now enumerate $S + B$ by interlacing $(0,x_i)$ and $(1,y_i)$. Do you see how to use this enumeration of $S + B$ to construct a bijection between $S + B$ and $S$? If so, then the rest of $A + B$ can be handled by the mapping $(0,x) \mapsto x$.

If $+$ were replaced by $\cup$ then you would need to worry about $S$ and $B$ having nonempty intersection, but this can also be handled by a slightly different argument. However, this argument works pretty much exactly as written when $B$ is finite rather than countably infinite.

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