Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 2

Proof
Let $H = B \cup \displaystyle \bigcup \mathcal A$

that $H$ is not connected.

That is, that $H$ is disconnected.

We have that:
 * $\forall C \in \mathcal A: B \cap C \ne \varnothing$

Thus:
 * $\exists x \in H: x \in B \cap C$

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 $\left\langle{A_\alpha}\right\rangle$.

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

As $s \in V$, it follows that the $\left\langle{A_\alpha}\right\rangle$ 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 $T$.