Suppose \(B_\alpha\subseteq X\) are connected for every \(\alpha\in I\) and there is a point \(p\in X\) such that \(p\in B_\alpha\) for every \(\alpha\in I\). Then \(\cup_{\alpha\in I} B_\alpha\) is connected.
Proof
For the sake of contradiction assume \(U\) and \(V\) disconnect \(\cup_{\alpha} B_\alpha\). Without loss of generality let \(p\in U\).
Since for a fixed \(\alpha\) the set \(B_\alpha\) is connected and \(B_\alpha\subseteq U\cup V\) (0x680b53b0)
implies that \(B_\alpha\subseteq U\).
Therefore, \(\cup_{\alpha} B_\alpha =U\), which is a contradiction.