Jeremy Rickard's answer uses the fact that a symmetric group is not the union of a countable chain of proper subgroups. The following easy proof of that fact is quoted from Fred Galvin, *Generating countable sets of permutations*, J. London Math. Soc. (2) 51 (1995), 230-242.

[. . .] permutations are regarded as right operators, and are composed from left to right. [. . .] The *pointwise stabilizer* of a subset $X$ of $E$ is the group $S_X=\{\pi\in\operatorname{Sym}(X):X\subseteq\operatorname{fix}(\pi)\}$.

We shall make heavy use of the following lemma, which was proved by Dixon, Neumann and Thomas [**3**, Lemma, p. 580] for the case $|E|=\aleph_0$, and generalized by Macpherson and Neumann [**11**, Lemma 2.1] to arbitrary infinite sets.

LEMMA 2.1. *Let $E$ be an infinite set. If $E=A\cup B\cup C$ where $A,B,C$ are disjoint sets and $|A|=|B|=|C|$, then $\operatorname{Sym}(E)=S_AS_BS_A\cup S_BS_AS_B$.*

*Proof.* Let $\kappa=|E|$. Consider a permutation $\pi\in\operatorname{Sym}(E)$. It is easy to see that $\pi\in S_AS_BS_A$ if (and only if) $|(B\cup C)\setminus A\pi^{-1}|=\kappa$. In particular, $\pi\in S_AS_BS_A$ if $|C\setminus A\pi^{-1}|=\kappa$; similarly, $\pi\in S_BS_AS_B$ if $|C\setminus B\pi^{-1}|=\kappa$. At least one of these alternatives must hold, since $C=(C\setminus A\pi^{-1})\cup(C\setminus B\pi^{-1})$.

[. . .]

THEOREM 3.1. *Let $E$ be an infinite set. Every countable subset of $\operatorname{Sym}(E)$ is contained in a*

$4$-generator subgroup of $\operatorname{Sym}(E)$.

*Proof.* We may assume that $E=\mathbb Z\times\mathbb Z\times T$, where $|T|=|E|=\kappa$. Let $E_0=\{0\}\times\{0\}\times T$. Choose $A\subset E_0$ with $|A|=|E_0\setminus A|=\kappa$; let $C=E_0\setminus A$ and let $B=E\setminus E_0$. Choose an involution $\varepsilon\in\operatorname{Sym}(E)$ so that $B\varepsilon=A$. Define $\alpha,\delta\in\operatorname{Sym}(E)$ by setting $(m,n,t)\alpha=(m+1,n,t),(0,n,t)\delta=(0,n+1,t)$, and $(m,n,t)\delta=(m,n,t)$ for $m\ne0$.

Let a countable set $H\subseteq\operatorname{Sym}(E)$ be given; we shall show that $H\subseteq\langle\alpha,\beta,\delta,\varepsilon\rangle$ for some $\beta\in\operatorname{Sym}(E)$. By Lemma 2.1, we may assume that $H\subseteq S_A\cup S_B$. Let $H'=(H\cap S_B)\cup\varepsilon(H\cap S_A)\varepsilon$. Then $H'$ is a countable subset of $S_B$; let $H'=\{\phi_i:i\in\mathbb Z\}$. Since $\operatorname{supp}(\phi_i)\subseteq E_0$, we can define $\hat\phi_i\in\operatorname{Sym}(T)$ so that $(0,0,t)\phi_i=(0,0,t\hat\phi_i)$ for $t\in T,i\in\mathbb Z$. Finally, define $\beta\in\operatorname{Sym}(E)$ by setting
$$(m,n,t)\beta=
\begin{cases}
(m,n,t\hat\phi_m)&\text{if }n\ge0,\\
(m,n,t)&\text{if }n\lt0.\\
\end{cases}$$
Then $\phi_i=(\alpha^i\beta\alpha^{-i})\delta^{-1}(\alpha^i\beta^{-1}\alpha^{-i})\delta$ for each $i\in\mathbb Z$; thus we have $H'\subseteq\langle\alpha,\beta,\delta\rangle$ and $H\subseteq H'\cup\varepsilon H'\varepsilon\subseteq\langle\alpha,\beta,\delta,\varepsilon\rangle$.

COROLLARY 3.2. *A symmetric group is not the union of a countable chain of proper subgroups.*

[. . .]

THEOREM 3.3. *Let $E$ be an infinite set. Every countable subset of $\operatorname{Sym}(E)$ is contained in a*

$2$-generator subgroup of $\operatorname{Sym}(E)$.

*Proof.* [. . .]