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?



Does the symmetric group on an infinite set have a minimal generating set?
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.)


