How to show the ascending chain condition 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.
 A: Note that the increasing union of chains is a chain. Formally:
Suppose for $i\in\mathbb N$, $Q_i$ is a chain in $(A,\le)$ and for $i < j$ we have $Q_i\subseteq Q_j$, then $\bigcup Q_i=Q$ is a chain.
Proof: Given $a,b\in Q$ then for some $i,j$ we have $a\in Q_i$ and $b\in Q_j$. Since either $Q_i\subseteq Q_j$ or $Q_j\subseteq Q_i$ we have that $a,b\in Q_{\max\{i,j\}}$ and therefore $a\le b$ or $b\le a$. Thus, $Q$ is a chain.
By the same argument you can also show that if $a$ is the minimum of $Q_i$ and $b$ is the maximum of $Q_i$, for all $i\in\mathbb N$ then $a,b$ are also the minimum/maximum of $Q$.

Now if you have an infinite chain of strict inclusions then there is some $a_i\in Q_i\setminus\bigcup_{j<i}Q_j$ such that $a<a_i<b$ (simply because there is a strict inclusion between the $Q_i$'s).
The collection $\{a_i\mid i\in\mathbb N\}\subseteq Q=\bigcup Q_i$, and $a_i\neq a_j$ since if $j<i$ then $a_i\notin Q_j$, but $a_j\in Q_i$. This means that the chain $Q$ is infinite, and its minimum is $a$ and its maximum is $b$ - which contradicts the property of every chain strictly between $a$ and $b$ is finite.
