Union of Overlapping Convex Sets in Toset is Convex/Infinite Union

Theorem
Let $\left({ S, \preceq }\right)$ be a totally ordered set.

Let $\mathcal A$ be a set of convex subsets of $S$.

Suppose that for any $P, Q \in \mathcal A$, there are elements $C_0, \dots, C_n \in \mathcal A$ such that:
 * $C_0 = P$
 * $C_n = Q$
 * For $k=0, \dots, n-1$: $C_k \cap C_{k+1} \ne \varnothing$

Then $\bigcup \mathcal A$ is convex in $S$.

Proof
Let $a, c \in \bigcup \mathcal A$.

Let $b \in S$.

Let $a \prec b \prec c$.

Since $a, c \in \bigcup \mathcal A$, there are $P, Q \in \mathcal A$ such that $a \in P$ and $c \in Q$.

By the premise, there are elements $C_0, \dots, C_n \in \mathcal A$ such that:
 * $C_0 = P$
 * $C_n = Q$
 * For $k=0, \dots, n-1$: $C_k \cap C_{k+1} \ne \varnothing$

Applying Union of Overlapping Intervals is Interval inductively:
 * $\displaystyle \bigcup_{k=0}^n C_k$ is convex.

Since $a,c \in \displaystyle \bigcup_{k=0}^n C_k$, by the definition of convexity:
 * $b \in \displaystyle \bigcup_{k=0}^n C_k$

Thus $b \in \bigcup \mathcal A$.

Since this holds for all such triples $a,b,c$, $\mathcal A$ is convex.