I am trying to understand the proof that the following holds in ZF: if there is an injection from $\omega$ to $\mathcal{P}(X)$, then there is a surjection from $X$ to $\omega$.
I am reading the proof in Herrlich's Axiom of Choice, p.47.
I understand why the proof boils down to recursively defining $g: \omega \to \mathcal{P}(X)$, such that the $g(n)$'s are non-empty and pairwise disjoint.
I transcribe the proof below adding markers (1) and (2) for the points that are unclear to me.
(1) How can we be sure $n^\star$ exists?
(2) Is this supposed to be $g(n)=(X\setminus f(n^\star))\setminus \cup_{m<n} g(m)$? If not, how can we be sure we satisfy the pairwise disjoint requirement?
Let $f: \omega \to \mathcal{P}(X)$ be an injection. Define recursively a map $g: \omega \to \mathcal{P}(X)$ such that the $g(n)$'s are non void and pairwise disjoint: For $n\in \omega$, assume that the $g(m)$'s are defined for all $m<n$ such that the set $\{ f(k) \setminus \cup_{m<n} g(m): k \geq n \}$ is infinite.
Define (1)
$n^\star = min \{ k \, | \, k \geq n \text{ and } f(k) \setminus \cup_{m<n} g(m) \neq \emptyset \neq \big(X \setminus f(k)\big) \setminus \cup_{m<n} g(m) \}$
In case $\{ f(k) \setminus \big(f(n^\star) \bigcup \cup_{m<n} g(m)\big) \, | \, k \geq n^\star \} $ is infinite, define $g(n)=f(n^\star) \setminus \cup_{m<n} g(m)$.
Otherwise, define (2) $g(n)=X \setminus \big( f(n^\star) \setminus \cup_{m<n} g(m) \big).$