Union of Connected Sets with Non-Null Intersection is Connected

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

Let $H = \bigcup A_\alpha \cup B$ where:
 * $(1): \quad \left\{{A_\alpha}\right\}$ is a set of subsets of $S$ where $\alpha$ is the element of some indexing set
 * $(2): \quad B \subseteq S$
 * $(3): \quad$ All of $A_\alpha$ and $B$ are connected
 * $(4): \quad \forall \alpha: A_\alpha \cap B \ne \varnothing$

Then $H$ is connected.

Proof
that $H$ is not connected.

That is, that $H$ is disconnected.

From $(4)$:
 * $\exists x \in H: x \in A_\alpha \cap B$

From the definition of disconnected, there exist separated sets $U, V$ whose union is $H$.

So:
 * $\forall s \in S: s \in U \lor s \in V$

$s \in V$.

Then $H, V$ serve as separated sets whose union is a cover for $\left\{{A_\alpha}\right\}$.

Note that the disconnection is valid as:
 * $\exists x \in U \cap \left\{{A_\alpha}\right\}$

As $s \in V$, it follows that the $\left\{{A_\alpha}\right\}$ are disconnected.

From this contradiction it follows that $s \notin V$.

Hence $s \in U$.

This holds for arbitrary $s \in H$.

Hence: $H \subseteq U$

But as $U$ and $V$ are separated, $U \cap H$ is empty, as required per the definition of a disconnected set.

This contradicts our deduction that $H \subseteq U$.

Hence the sets $U, V$ are not separated sets in $H$.

Thus $H$ is connected in $T$.