To clarify the terms in the question above:
The symmetric group Sym($\Omega$) on a set $\Omega$ consists of all bijections from $\Omega$ to $\Omega$ under composition of functions. A generating set $X \subseteq \Omega$ is minimal if no proper subset of $X$ generates Sym($\Omega$).
This might be a difficult question, but perhaps the answer is known already?
I think it follows from Theorem 1.1 of "Subgroups of Infinite Symmetric Groups" by Macpherson and Neumann (J. London Math. Soc. (1990) s2-42 (1): 64-84) that there is no minimal generating set of $S(\Omega)$for infinite $\Omega$.
The theorem states that any chain of proper subgroups of $S(\Omega)$ whose union is $S(\Omega)$ must have cardinality strictly greater than $|\Omega|$.
Now suppose $X$ is a minimal generating set. Let $C=\{x_0,x_1,\dots\}$ be a countable subset of $X$. If $$H_i=\langle X\setminus C,x_0,\dots,x_i\rangle$$ for $i\in\mathbb{N}$, then $H_0<H_1<\dots$ is a countable chain of proper subgroups whose union is $S(\Omega)$, contradicting the theorem.
(Note: There's a paper of Bigelow pointing out some unstated set-theoretic assumptions in Macpherson and Neumann's paper, but I don't think that affects the theorem I mention.)
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. [. . . .]
The canonical paper on the subject (don't be fooled by the publication date, it had been around for years before then) is George Bergman's gem:
Bergman, George M., Generating infinite symmetric groups., Bull. Lond. Math. Soc. 38, No. 3, 429-440 (2006). ZBL1103.20003.
