MHB Countable Dense Open Sets in Compact Hausdorff Spaces?

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In a compact Hausdorff space, any countable collection of dense open sets results in a dense intersection. The problem of the week (POTW) invites participants to demonstrate this property. A correct solution has been provided by Opalg, which is available for review. The discussion emphasizes the significance of this result in topology. Engaging with such problems enhances understanding of compactness and density in topological spaces.
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Here is this week's POTW:

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Show that in a compact Hausdorff space, any countable collection of dense open sets has dense intersection.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to Opalg for his correct solution, which you can read below:
This is one form of the Baire category theorem. The proof relies on the fact that a compact Hausdorff space $X$ is regular, meaning that if $U$ is an open subset of $X$ and $x\in U$ then there is an open subset $V$ such that $x\in V\subseteq \overline{V} \subseteq U$, where $ \overline{V}$ is the closure of $V$.

Suppose that $\{U_n\}_{n\in\Bbb N}$ is a countable collection of dense open subsets of $X$. Let $W$ be an open subset of $X$. Then we want to show that $$W\cap \bigcap_{n\in\Bbb N}U_n \ne\emptyset$$.

Since $U_1$ is dense in $X$, $W\cap U_1\ne\emptyset$. So there exists a nonempty open set $V_1$ such that $\overline V_1 \subseteq W\cap U_1$.

Since $U_2$ is dense in $X$, $U_2\cap V_1\ne\emptyset$. So there exists a nonempty open set $V_2$ such that $\overline V_2 \subseteq V_1\cap U_2$.

Continuing in this way, we get a decreasing sequence of nonempty open sets $\{V_n\}_{n\in\Bbb N}$ such that $$\overline V_n \subseteq W\cap \bigcap_{k=1}^n U_k$$. The sets $\overline V_n$ are compact and satisfy the finite intersection property. So the intersection $$\bigcap_{n\in\Bbb N}\overline V_n$$ is nonempty and contained in $$W\cap \bigcap_{n\in\Bbb N}U_n$$. That completes the proof.