It is sufficient to show for one set only, so we show that $Y\cup A$ is connected. Suppose it is not, then let $Y\cup A=C\cup D$ where $C$ and $D$ are nonempty disjoint open subsets of $Y\cup A$ . Since $Y$ is a connected subset of $Y\cup A$ , it must lie within either $C$ or $D$ (Lemma 23.2), so suppose $Y\subset C$ , so that $D\subset A$ . We have $X=C\cup D\cup B$ . Using Lemma 23.1, no limit point of $C$ can be in $D$ , and no limit point of $B$ can be in $D\subset A$ , so that $B\cup C$ is closed, and $D$ is open in $X$ . But no limit point of $D\subset A$ can lie in $C$ or $B$ . So that $D$ is closed in $X$ . Therefore, $D$ is both open and closed in $X$ . Contradiction.