An extension of Baire's category theorem

This is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.

Let's assume $X$ is a Baire space. Suppose $(S_n)_{n\in \mathbb{N}}$ is a sequence of open sets in $X$. We always have $\text{int}(\text{cl}(\bigcap_{n=1}^\infty S_n))\subseteq \text{int}(\bigcap_{n=1}^\infty \text{cl}(S_n))$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $U\subseteq \bigcap_{n=1}^\infty \text{cl}(S_n)$, we have $U\subseteq \text{cl}(\bigcap_{n=1}^\infty S_n)$.

So fix some $U$. Note that an open subspace of a Baire space is Baire. [Why? If $(V_n)_{n\in\mathbb{N}}$ is a sequence of dense open subsets of $U$, then $(V_n\cup (X\setminus \text{cl}(U)))_{n\in\mathbb{N}}$ is a sequence of dense open subsets of $X$. The intersection of these sets is $(\bigcap_{n=1}^\infty V_n)\cup (X\setminus \text{cl}(U))$, and the fact that this is dense in $X$ implies that $(\bigcap_{n=1}^\infty V_n)$ is dense in $U$.]

So defining $V_n = U\cap S_n$, we have that each $V_n$ is open in $U$, and since $U\subseteq \text{cl}(S_n)$ for all $n$, each $V_n$ is dense in $U$. Since $U$ is Baire, $\bigcap_{n=1}^\infty V_n$ is dense in $U$, which implies $U\subseteq \text{cl}(\bigcap_{n=0}^\infty S_n)$, as was to be shown.

Here is my attempt to prove it. It is inspired by the proof of [Zălinescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.

Theorem. Let $X$ be a locally compact Hausdorff space, and $\{S_n\}$ be a sequence of open sets in X. Then $\mathrm{cl}(\cap_{n=1}^\infty S_n)$ and $\cap_{n=1}^\infty \mathrm{cl}(S_n)$ have the same interior.

proof. It suffices to show that $\mathrm{int}(\cap_{n=1}^\infty \mathrm{cl}(S_n))\subset \mathrm{cl}(\cap_{n=1}^\infty S_n)$. To this end, we only need to prove for any given nonempty open set $U \subset \cap_{n=1}^\infty \mathrm{cl}(S_n)$ that \begin{equation} \label{eq:usnonempty} U\cap(\cap_{n=1}^\infty S_n) \neq \emptyset. \end{equation} In the sequel, we will define a sequence of compact sets $\{C_n\}_{n=0}^\infty$ such that $C_n$ has nonempty interior and \begin{equation} C_{n+1} \subset C_{n} \subset U\cap(\cap_{k=1}^n S_k) \quad \text{ for each } \quad n \ge 0, \end{equation} where $\cap_{k=1}^0 S_k=X$. Once this is done, Cantor's Theorem will yield \begin{equation} \label{eq:nestnonempty} \emptyset \;\neq \;\cap_{n=1}^\infty C_n \;\subset\; U\cap (\cap_{n=1}^\infty S_n), \end{equation} which gives us what we want.

We define $\{C_n\}$ inductively. As $U$ is a nonempty open set and $X$ is locally compact, we can take a compact set $C_0\subset U$ such that $\mathrm{int}(C_0)\neq \emptyset$. Assume that $C_n$ is already defined for an $n\ge 0$ so that $C_n\subset U\cap(\cap_{k=1}^n S_k)$ and $\mathrm{int}(C_n)\neq \emptyset$. Recalling that $U\subset\mathrm{cl}(S_{n+1})$, we have $\mathrm{int}(C_n)\subset \mathrm{cl}(S_{n+1})$, which implies that $\mathrm{int}(C_n) \cap S_{n+1} \neq\emptyset$. Since $\mathrm{int}(C_n)\cap S_{n+1}$ is open, invoking again the local compactness of $X$, we can take a compact set $C_{n+1}\subset \mathrm{int}(C_n) \cap S_{n+1}$ such that $\mathrm{int}(C_{n+1})\neq \emptyset$. Clearly, $C_{n+1}\subset C_n$ and $C_{n+1}\subset C_{n}\cap S_{n+1} \subset U\cap(\cap_{k=1}^{n+1} S_k)$. This finishes the induction and completes the proof.