Set of Chains is Closed under Chain Unions under Subset Relation
Theorem
Let $\struct {P, \le}$ be an ordered set.
Let $\map {\operatorname {ch} } P$ be the set of chains in $P$.
Then $\map {\operatorname {ch} } P$ is closed under chain unions.
That is, if $\CC$ is a chain in $\map {\operatorname {ch} } P$ with respect to the subset ordering, the union $\ds \bigcup \CC$ of $\CC$ is also a chain in $P$.
Proof
Let $\ds D = \bigcup \CC$.
Observe that any $C \in \CC$ is a chain in $P$, hence $C \subseteq P$.
Therefore, $D \subseteq P$, by Set Union Preserves Subsets.
Now suppose that $a, b \in D$.
By definition of union, there exist $A, B \in \CC$ such that $a \in A$ and $b \in B$.
Since $\CC$ is a chain in $\struct {\map {\operatorname {ch} } P, \subseteq}$, either $A \subseteq B$ or $B \subseteq A$.
Hence either $a \in B$ or $b \in A$.
Both $A$ and $B$ are chains in $P$, and thence $a \le b$ or $b \le a$.
Therefore, $D$ is totally ordered by $\le$.
That is, it is a chain in $P$.
$\blacksquare$