I have been having trouble with an exercise from Advanced Linear Algebra by Steven Roman, Chapter 5 exercise 6.
Show that an $R$-module $M$ satisfies the ACC (ascending chain condition) for submodules if and only if the following condition holds. Every nonempty collection $\mathcal{S}$ of submodules of $M$ has a maximal element. That is, for every nonempty collection $\mathcal{S}$ of submodules of $M$ there is an $S \in \mathcal{S}$ with the property that $T \in \mathcal{S} \implies T \subseteq S$.
If the condition holds, the $M$ obvious satisfies the ACC because for an ascending chain $T_1 \subseteq T_2 \subseteq T_3 \dots$ we can take the sequence as a collection of submodules $\mathcal{S}$ which by condition has a maximal element $S$. This implies that the sequence stabilizes at $S$.
However, what is confusing me is the "if and only if". Every vector space is trivially a module, so if we look at $\mathbb{R}^2$ as an $\mathbb{R}$-module, and take the collection of one-dimensional subspaces (which are also submodules), that collection doesn't satisfiy the maximal element condition, but $\mathbb{R}^2$ (as every finite dimensional vector space) does satisfy the ascending chain condition. This can be easily seen from the dimensions of the subspaces, $T_1 \subseteq T_2 \subseteq T_3 \dots \implies 0 \leq \text{dim} \; T_1 \leq \text{dim} \; T_2 \leq \text{dim} \; T_3 \leq \dots \leq 2,$ from which it follows that every such sequence does stabilize.
This is an incorrect definition of "maximal element," and you are right that your example disproves it.
The correct definition is something more like this:
That is: nothing in $\mathcal S$ properly contains $S$. With that correction, all of the one-dimensional subspaces in $\mathbb R^2$ are maximal with respect to each other.
Now, as for the relation to the converse: Zorn's lemma says that if the ACC is satisfied like this for chains, then any nonempty subset has a maximal element. (With the 'correct' definition of maximal.)