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

Since $A$ is at most countable there is a bijection $\phi :A\to N_0$ 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 $\phi$ to $B$ gives a bijection of $B$ onto $S:=\phi [B]\subseteq \mathbb{N}$, so $B\approx S$. It therefore suffices to prove that every subset of $\mathbb{N}$ is at most countable, since then $S\approx n^{\mathrm{\prime }}$ 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 $T\subseteq 0=\mathrm{\varnothing }$ 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 $m\mathrm{\notin }T$ then $T=T\setminus \{m\}\approx k$. If $m\in T$, extend $u$ by $u(m):=k$; since $k\mathrm{\notin }k$ 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 $S\mathrm{\ne }\mathrm{\varnothing }$, 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 $S\mathrm{\ne }\mathrm{\varnothing }$) 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 $m\mathrm{\ne }n$ 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 $j\mathrm{\notin }K$ 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.