Any neighbourhood of a closed decreasing sequence in a compact space must contain one of the sets: why is the space required also to be Hausdorff?

Question

$\newcommand{\o}{\mathcal{O}}$Apologies for the perhaps strangely phrased title - the character limit can be a pain.

Let $X$ be a compact Hausdorff space and $\{F_n\subseteq X:n\in\Bbb N\}$ a decreasing sequence of nonempty closed sets in $X$. For any neighbourhood $\o$ of their intersection $\bigcap_{n\in\Bbb N}F_n$, show that there must exist an $N\in\Bbb N$ such that for all $n\ge N$, $F_n\subseteq\o$.

I have done this exercise - but my proof nowhere used the Hausdorff property. I ask: why is it needed?

My solution:

As the $F_n$ have the finite intersection property, and the space is compact, their intersection is nonempty. $\o$ is then nonempty, and wlog may be considered open. Consider then the family of sets $G_n=F_n\cap X\setminus\o$. For all $n$, $G_n$ is the intersection of two closed sets and thus closed. As $F_{n+1}\subseteq F_n,\,G_{n+1}\subseteq G_n$ for all $n$. Assume for the sake of contradiction that $F_n$ is not a subset of $\o$ for any $n$; this is equivalent to the assumption that $G_n$ is nonempty for all $n$.

If none of the $G_n$ are empty, then $\{G_n:n\in\Bbb N\}$ is a decreasing sequence of closed nonempty sets and thus has the finite intersection property, and thus has nonempty intersection due to the compactness of $X$. Then $\bigcap_{n\in\Bbb N}F_n\cap X\setminus\o\neq\varnothing$, which implies the existence of $x\in\bigcap_{n\in\Bbb N}F_n$ such that $x\notin\o$, which contradicts the assertion that $\o$ is a neighbourhood of $\bigcap_{n\in\Bbb N}$. Since $\{G_n\}$ will always have nonempty intersection unless there is an $N\in\Bbb N$ such that $G_N=\varnothing$, this shows that such an $N$ must exist and that $F_N\cap X\setminus\o=\varnothing,\,F_N\subseteq\o$. It follows from the decreasing property that $\forall n\ge N,\,F_n\subseteq\o\quad\blacksquare$

This exercise was left by Royden in the fourth edition of their text "Real Analysis". My proof nowhere requires the Hausdorff assertion (as far as I can tell) so I'm left wondering if Royden made an oversight. Is this the case?

Answer 1

I think you're right. The statement

Let $F_n$ be a FIP family of non-empty closed sets of $X$. If $O$ is open and $\bigcap_n F_n \subseteq O$ then for some $N$: $F_N \subseteq O$.

only needs $X$ to be compact (to apply the FIP criterion for compactness) and doesn't need separation axioms. (and if $n \ge N$ by decreasingness $F_n \subseteq O$ as well etc).

Indeed, if the conclusion failed, $\{F_n\setminus O\mid n \in \Bbb N\}$ would be a FIP family of non-empty closed subsets of $X$ with empty intersection, which cannot be.