Equivalence of Forms of Axiom of Countable Choice

Theorem
The following forms of the Axiom of Countable Choice are equivalent in $\mathrm{ZF}^-$:

Form 1 implies Form 2
Suppose that Form 1 holds.

Let $S$ be a countable set of non-empty sets.

Then $S$ is either finite or countably infinite.

If $S$ is finite, then it has a choice function by the Principle of Finite Choice.

Suppose instead that $S$ is countably infinite.

Then there is a bijection $f: \N \to S$.

Thus by Form 1, there is a mapping $g: \N \to \bigcup S$ such that for all $n \in \N$, $g(n) \in f(n)$.

Then $g \circ f^{-1}: S \to \bigcup S$ is a choice function for $S$.

As every countable set has a choice function, Form 2 holds.

Form 2 implies Form 1
Suppose that Form 2 holds.

Let $\langle x_n \rangle_{n \in \N}$ be a sequence of non-empty sets.

By the Axiom of Replacement, $I = \{ x_n: n \in \N \}$ is a set.

By Surjection from Natural Numbers iff Countable, $I$ is countable.

By Form 2, $I$ has a choice function $h: I \to \bigcup I$.

Define a sequence in $\bigcup I$ by letting $y_n = h(x_n)$.

Then by the definition of choice function, for each $n \in \N$, $y_n \in x_n$.

As this holds for all such sequences, Form 1 holds.