Union of Connected Sets with Common Point is Connected/Proof 1

Proof
Let $U, V$ be disjoint open sets of the subspace of $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ such that:
 * $U \cup V = \displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$

assume $x \in U$.

From Set is Subset of Union:
 * $\displaystyle \forall \alpha \in A : B_\alpha \subseteq \bigcup_{\alpha \mathop \in A} B_\alpha = U \cup V$

From Connected Subset of Union of Disjoint Open Sets:
 * $\forall \alpha \in A : B_\alpha \subseteq U$

From Union is Smallest Superset:
 * $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha = U$

Hence $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ admits no separation.

It follows that $\displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$ is a connected set of $T$ by definition.