Every subset of an at most countable set is at most countable

Since $A$ is at most countable there is a bijection $\varphi : A \to N_0$ where $N_0 = n$ for some $n \in \mathbb{N}$ or $N_0 = \mathbb{N}$; in either case $N_0 \subseteq \mathbb{N}$, and restricting $\varphi$ to $B$ gives a bijection of $B$ onto $S := \varphi[B] \subseteq \mathbb{N}$, so $B \approx S$. It therefore suffices to prove that every subset of $\mathbb{N}$ is at most countable, since then $S \approx n'$ or $S \approx \mathbb{N}$ and transitivity carries the conclusion back to $B$.

Every subset of a natural number is finite: by strong induction on $n$, assume every subset of every $m < n$ is finite. If $n = 0$ then a subset $T \subseteq 0 = \varnothing$ is empty and $T \approx 0$. Otherwise $n = \sigma(m)$ by [L7], with $m < n$; given $T \subseteq \sigma(m) = m \cup \{m\}$, the set $T \setminus \{m\}$ is a subset of $m$, so the hypothesis at $m$ gives a bijection $u : T \setminus \{m\} \to k$ for some $k \in \mathbb{N}$. If $m \notin T$ then $T = T \setminus \{m\} \approx k$. If $m \in T$, extend $u$ by $u(m) := k$; since $k \notin k$ by irreflexivity of membership, the value $k$ is not already taken and the extension is a bijection $T \to k \cup \{k\} = \sigma(k)$. In both cases $T$ is finite, so the claim holds for $n$ and hence for all $n$.

Case $S$ bounded: assume there is $b_0 \in \mathbb{N}$ with $s \le b_0$ for every $s \in S$. Then $s < \sigma(b_0)$ for every $s \in S$ by [L6], that is, $S \subseteq \sigma(b_0)$.

Case $S$ unbounded: assume that for every $b \in \mathbb{N}$ there is $s \in S$ with $b < s$. Then $S \ne \varnothing$, and for each $s \in S$ the set $\{\, t \in S : s < t \,\}$ is nonempty, so [L3] makes $\nu(s) := \min \{\, t \in S : s < t \,\}$ a well-defined element of $S$ with $s < \nu(s)$; this defines a function $\nu : S \to S$ with no arbitrary choices.

In the bounded case $S$ is a subset of the natural number $\sigma(b_0)$, hence finite by step 1.2, hence at most countable.

In the unbounded case apply [L5] with $X = S$, $x_0 = \min S$ (available by [L3] since $S \ne \varnothing$) and $F = \nu$: there is $e : \mathbb{N} \to S$ with $e(0) = \min S$ and $e(\sigma(n)) = \nu(e(n)) = \min \{\, t \in S : e(n) < t \,\}$ for every $n$.

For every $n$, $e(n) < e(\sigma(n))$ by the defining property of $\nu$; consequently $m < n$ implies $e(m) < e(n)$, by strong induction on $n$ (for $n = \sigma(j)$ and $m < n$ one has $m \le j$ by [L6], so either $m = j$, giving $e(m) < e(\sigma(j))$ directly, or $m < j$, giving $e(m) < e(j) < e(\sigma(j))$ by the hypothesis at $j$ and transitivity). Hence $e$ is injective: if $m \ne n$ then $m < n$ or $n < m$ by comparability, and irreflexivity forbids $e(m) = e(n)$.

For every $k$, $k \le e(k)$: again by strong induction, at $k = 0$ this is immediate, and for $k = \sigma(j)$ the hypothesis at $j$ gives $j \le e(j) < e(\sigma(j))$, so $j < e(k)$ and therefore $\sigma(j) \le e(k)$ by [L6], that is $k \le e(k)$.

$e$ is surjective onto $S$: let $t \in S$. The set $K = \{\, k \in \mathbb{N} : t \le e(k) \,\}$ contains $t$ by step 3.2, so $k^\ast := \min K$ exists by [L3]. If $k^\ast = 0$ then $e(0) = \min S \le t$ because $t \in S$, and $t \le e(0)$, so $e(0) = t$. Otherwise $k^\ast = \sigma(j)$ by [L7], and $j \notin K$ by minimality, so $e(j) < t$; then $t$ belongs to $\{\, u \in S : e(j) < u \,\}$, whence $e(k^\ast) = \min \{\, u \in S : e(j) < u \,\} \le t$, and with $t \le e(k^\ast)$ this gives $e(k^\ast) = t$. In both cases $t$ is a value of $e$.

In the unbounded case $e : \mathbb{N} \to S$ is therefore a bijection, so $S \approx \mathbb{N}$ and $S$ is countably infinite, hence at most countable.

Every $S \subseteq \mathbb{N}$ is either bounded above or not, so steps 2.1 and 5.1 cover all cases and every subset of $\mathbb{N}$ is at most countable; by the reduction of step 1.1 the subset $B$ of the at most countable set $A$ is at most countable.