Why is axiom of choice needed? (Equivalent conditions for Noetherian)

Let $R$ be a commutative ring with $1$ and let $M$ be a left $R$-module. On page 458 of Dummit and Foote's Algebra, 3rd edition, they show that $M$ is Noetherian (i.e. satisfies A.C.C. on submodules) if and only if every nonempty set of submodules of $M$ has a maximal element under inclusion. I don't understand why the axiom of choice comes into their proof, outlined below.

Their proof is as follows: Let $\Sigma$ be a nonempty set of submodules of $M$, and assume for a contradiction that $\Sigma$ does not have a maximal element under incusion. and take $M_1 \in \Sigma$. Then as $M_1$ is not maximal, there is some $M_2 \in \Sigma \setminus \{M_1\}$ such that $M_1 \subsetneq M_2$. We continue this way to get $M_1 \subsetneq M_2 \subsetneq M_3 \subsetneq \cdots$, contradicting the A.C.C. on submodules.

Where do they use the axiom of choice?

Index the elements of $\Sigma$ with elements of $I$. By assuming no element of $\Sigma $ is maximal, you assert that each set $\Sigma_i:=\{N\subset M\mid N\supset M_i\}$ is nonempty, and then you need a choice function $C:\{\Sigma_i\mid i\in I\}\to \Sigma$ to legitimately furnish infinitely many witnesses to the nonmaximality of each thing in $\Sigma$ all at once.

Once you have those witnesses, you inductively construct the countable ascending chain you mentioned. Without the choice function, it's unclear "how to choose a sock" in each $\Sigma_i$.