Let $X$ be a compact and Hausdorff space.
Let $\{U_n\}$ be a countable collection of subsets that are open and dense in X.
Let $O$ be an open set. Since $U_1$ is dense, it must intersect $U_1$ in a non-empty open set $O_1$. Let $x_1 \in O_1$ , and since the space is Hausdorff and compact exists $B_1(x_1)$ a open set that contains $x_1$ and $\overline{B_{1}(x_1)}$ is contained in $O_1$. Now, because $U_2$ is dense, intersects $B_{1}(x_1)$ in a non-empty open set $O_2$. In the same way, let $x_2 \in O_2$ , and take $B_2(x_2)$ such that the closure $\overline{B_{2}(x_2)}$ is contained in $O_2$.
With this process we obtained a nested sequence of non-empty closed sets $\overline{B_1} \supseteq \overline{B_2} \supseteq \overline{B_3} \supseteq ... \supseteq \overline{B_n} \supseteq ... $ . As $X$ is compact, then exists $ x \in \bigcap\limits_{n=1}^{\infty} \overline{B_n} $. In this way, $x \in O_n$, for each $n$ and hence $x \in O \cap \bigcap\limits_{n=1}^{\infty} {U_n}$. Thus $\bigcap\limits_{n=1}^{\infty} {U_n}$ intersects each non-empty set $O$ in a least one point, that is precisely, that $\bigcap\limits_{n=1}^{\infty} {U_n}$ is dense in $X$.