Let $U$ be an open subset of compact Hausdorff space $X$. Choose a non-empty subset $U_1$ in $X$ with $\overline{U}_1\subseteq U$. Repeating this construction without end, we can then find a sequence of open non-empty sets $\{U_n\}_{n\in\mathbb{N}}$ in $X$ with $U_{n+1}\subseteq \overline{U}_{n+1}\subseteq U_n$.
$X$ is a compact Hausdorff space but it may be not metrizable. Can I say that $\bigcap _{n\in\mathbb{N}}U_n\neq \emptyset$?
Pease help me to know it.
Yes, we have $D = \bigcap _{n\in\mathbb{N}}U_n \ne \emptyset$.
In fact, since $U_{n+1}\subset \overline{U}_{n+1}\subset U_n$, we get $$D = \bigcap _{n\in\mathbb{N}} \overline{U}_n .$$ The family $\{ \overline{U}_n \}$ clearly has the finite intersection property which means that each finite intersection of members is non-empty. It is well-known that a space $X$ is compact iff any family of closed subsets which has the finite intersection property has a non-empty intersection.