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.