Union of Connected Sets with Non-Empty Intersections is Connected

Theorem
Let $T$ be a topological space.

Let $\mathcal A$ be a set of connected subspaces of $T$.

Suppose that, for all $B, C \in \mathcal A$, the intersection $B \cap C$ is nonempty.

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

Corollary
Let $T$ be a topological space.

Let $\mathcal A$ be a set of connected subspaces of $T$.

Suppose there is a connected subspace of $T$ such that $B \cap C \ne \varnothing$ for all $C \in \mathcal A$.

Then $B \cup \bigcup \mathcal A$ is connected.

Proof
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 $B, C \in \mathcal A$, we have $B \cap C \ne \varnothing$, and hence $\epsilon \left({B}\right) = \epsilon \left({C}\right)$.

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

Proof of Corollary
Let $C \in \mathcal A$. Then, applying the theorem to $B$ and $C$, the union $B \cup C$ is connected.

Thus the set $\tilde{\mathcal A} = \left\{{B \cup C: C \in \mathcal A}\right\}$ satisfies the conditions of the theorem, and the claim follows.