The axiom of infinity promises us an inductive set that contains $\varnothing$, but does not promise us a countable inductive set.
I'm wondering how much you can weaken the ability of some of the other axioms to create new sets out of uncountable sets (choice, separation, replacement, union) without losing the provability of the existence of a countably infinite set.
For the purposes of this question, the weakened version of the axiom of choice is the axiom of countable choice.
For separation and replacement, I impose the additional constraint that the parameter $A$ in the statements of the axioms below must be a countable set, giving countable separation and countable replacement, respectively.
Here is the ordinary axiom schema of specification; the axiom schema of countable specification imposes the additional constraint that $A$ is countable.
$$ \exists B \mathop. (\forall x \mathop. x \in B \leftrightarrow x \in A \land \varphi(\cdots, A)) $$
Likewise for replacement
$$ \exists B \forall x \in A \mathop. (\exists y \mathop. \varphi(\cdots, x, y) \to \exists z \in B \mathop. \varphi(\cdots, x, z)) $$
For union, I define countable union as the axiom that allows unary unions of countable sets only.
$$ \exists B \mathop. \forall c \mathop. (c \in B \leftrightarrow (\exists D \mathop. c \in D \land D \in A)) \; \text{is the axiom of union} $$
$$ \text{$A$ is countable} \to \exists B \mathop. \forall c \mathop. (c \in B \leftrightarrow (\exists D \mathop. c \in D \land D \in A)) \; \text{is the axiom of countable union} $$
I don't think that the remaining axioms, pairing, extensionality, powerset, and regularity will help make countably infinite sets out of uncountable ones.