Union of Connected Sets with Non-Null Intersection is Connected

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

Let $S = \bigcup A_\alpha \cup B$ where:
 * $\left\{{A_\alpha}\right\}$ is a set of subsets of $X$ where $\alpha$ is the element of some indexing set
 * $B \subseteq X$
 * All of $A_\alpha$ and $B$ are connected
 * $\forall \alpha: A_\alpha \cap B \ne \varnothing$

Then $S$ is connected.

Proof
As the intersection between the $\left\{{A_\alpha}\right\}$ and $B$ is non-empty, we know that there exists some $\exists x \in A_\alpha \cap B$.

Suppose that $S$ is not connected.

From the definition of connected, we see that there then exist some separated sets $U, V$ whose union is a cover for $S$.

So for all $\forall s \in S$ either $s \in U$ or $s \in V$.

If $s \in V$, then $S, V$ serve as separated sets whose union now covers $\left\{{A_\alpha}\right\}$.

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

Hence, if it was true that our arbitrary $s$ is in $V$, then we would have that the $\left\{{A_\alpha}\right\}$ are disconnected.

Therefore $s$ is in $U$.

However, as this holds for any arbitrary $s \in S$, we have that $S \subseteq U$.

As $U$ and $V$ are separated, $U \cap S$ is empty, as required per the definition of a disconnection.

Hence the sets $U, V$ do not disconnect $S$, and so $S$ is connected.