Set of Chains is Closed under Chain Unions under Subset Relation

Theorem
For a poset $\left( P, \leq \right)$, let $ch\left( P \right)$ be the set of chains in $P$.

The union of a chain in the poset $\left(ch \left( P \right), \subseteq \right)$ is a chain in $P$.

Proof
Let $D = \bigcup C$ where $C$ is a chain in $\left(ch \left( P \right), \subseteq \right)$, then $D \subseteq P$, since members of $C$ are subsets of $P$.

Suppose that $a$ and $b$ belong to $D$; then there are $A$ and $B$ in $C$ such that $a \in A$ and $b \in B$.

Since $C$ is a chain in $\left(ch \left( P \right), \subseteq \right)$, $A \subseteq B$ or $B \subseteq A$; suppose the latter.

Then both of $a$ and $b$ belong to $A$, which is a chain in $P$, so $a \leq b$ or $b \leq a$.

Therefore, $D$ is a chain in $P$.