Just follow the hint. $Y$ is closed. Therefore, both sets $C$ and $D$ are closed in $X$ . Since $X$ is compact, they are compact as well, and, by exercise 5, since $X$ is also Hausdorff, there are open and disjoint $U$ and $V$ containing them. $A-U-V$ are closed, nonempty (otherwise, since $U$ and $V$ are disjoint and both contain a point of $Y$ , there would be a separation of $A$ , which is connected) and simply ordered by proper inclusion (because the initial sequence is, otherwise the result would not hold), therefore, they satisfy the finite intersection property (why?), and since $X$ is compact, the intersection is nonempty. That implies that there is a point in $Y$ that is not in $U\cup V$ and, therefore, not in $C\cup D$ . Contradiction.