The Question
Let $A$ be a set. It is true that there exists some cardinality $\kappa_A$ such that whenever $\Sigma$ is a $\sigma$-algebra, and $\Sigma$ contains a generating set of cardinality $|A|$, then $|\Sigma| \leq \kappa_A$. The proofs I know all either involve transfinite induction up to $\omega_1$, or the full axiom of choice. Is there a way to prove this fact that doesn't use transfinite induction, and doesn't use the full axiom of choice (but may use the axiom of countable choice, or the axiom of dependent choice)? The cardinality $\kappa_A$ need not be optimal.
Background
I'm trying to construct the free abstract $\sigma$-algebra (i.e. not subsets of a set) generated by a set $A$ using simple tools, and in particular, without the full axiom of choice. To do this, it suffices to bound the size of any $\sigma$-algebra generated by $A$. However, I'm having trouble coming up with a simple explanation for why such a $\sigma$-algebra is bounded.
I've seen a proof that uses transfinite induction up to $\omega_1$ to effectively list out all elements of $\Sigma$, when $A$ generates $\Sigma$. This gives the bound $\kappa_A \leq \max(2^{\aleph_0}, |A|)$. While this doesn't use the full axiom of choice, I don't think it is simple to explain to someone who isn't already aware that $\omega_1$ can be well-ordered without the axiom of choice, or who isn't already familiar with transfinite induction. So I am looking for an alternative proof, if such a proof exists.