Union of Connected Sets with Non-Empty Intersections is Connected

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $I$ be an indexing set.

Let $\mathcal A = \left \langle{A_\alpha}\right \rangle_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.

Let $\mathcal A$ be such that no two of its elements are disjoint:
 * $\forall B, C \in \mathcal A: B \cap C \ne \varnothing$

Then $\displaystyle \bigcup \mathcal A$ is itself connected.

Proof
Let $A := \displaystyle \bigcup \mathcal A$.

Let $D = \left\{{0, 1}\right\}$, with the discrete topology.

Let $f: A \to D$ be continuous.

To show that $A$ is connected, we need to show that $f$ is not a surjection.

Since each $C \in \mathcal A$ is connected and the restriction $f \restriction_C$ is continuous:
 * $f \left({C}\right) = \left\{{\epsilon \left({C}\right)}\right\}$

where $\epsilon \left({C}\right) = 0$ or $1$.

But, for all $B, C \in \mathcal A$:
 * $B \cap C \ne \varnothing$

Hence $\epsilon \left({B}\right) = \epsilon \left({C}\right)$.

Thus $f$ is constant on $A$ as required.