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.

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.