Let $(A,\le)$ be a poset. Suppose that for any $a < b \in A$ and for any chain $Q$ of $A$ whose maximum and minimum are $a$ and $b$ respectively, $Q$ is finite. Let $C$ be the set of all chains of $A$, ordered under inclusion. I'd like to show that the poset $(C,\subset)$ satisfies the ascending chain condition (ACC).
I tried to show it by contradiction. Negating the ACC, we can say the existence of an infinite sequence $(Q_i)_{i\in\mathbb{N}}$ of elements of $C$ such that $Q_0 \subsetneq Q_1 \subsetneq Q_2 \subsetneq \cdots$. I think this contradicts the fact that all elements of $C$ are finite, but I cannot show it vigorously.
EDIT: What I really wanted to prove was that $(C, \subset)$ satisfies the ACC, where C is the set of chains whose whose maximum and minimum are $a$ and $b$ respectively. I'm sorry, but the answers seem to still hold.