Doubt about the proof of a property of compact subspaces of a union set with the union topology

Question

In p.153 of Algebraic Topology from a Homotopical Viewpoint we can find the following lemma:

5.1.10 Lemma. Let $X=\bigcup X_n$, $X_1\subset X_2\subset \ldots$ be a Hausdorff space with the union topology. Then every compact subset $K\subset X$ lies iniside $X_n$ for some $n$

Here is the proof:

Proof: If the conclusion were not so, then there would exist a sequence $\{x_n\}$ in $K$ satisfying $x_n\in X_n$. Now, any such sequence forms a closed subset of $X$, since its intersection with each $X_n$ is finite and hence closed in $X_n$. Here we are using the fact that $X$ is Hausdorff, implying that $X_n$ is also Hausdorff, so that points are closed in $X_n$. Therefore, the subsequences $\{x_m,x_{m+1},\ldots\}$, $m=1,2,3$ for a nested system of closed subsets of $K$whose intersection is empty, although the intersection of every finite subsystem is nonempty. And this would give us a contradiction to the compactness of $K$.

what is the property of Hausdorff spaces that is not satisfied in the reasoning?

Answer 1

I don't quite follow what is your question, but here is a comment, trying to clarify the proof. Thank you for supplying the proof, but it came with a couple of typos, so here is a corrected version (in particular, $x_n\not\in X_n$, rather than $x_n\in X_n$).

Proof: If the conclusion were not so, then there would exist a sequence $\{x_n\}$ in $K$ satisfying $x_n\not\in X_n$. Now, any such sequence forms a closed subset of $X$, since its intersection with each $X_n$ is finite and hence closed in $X_n$. Here we are using the fact that $X$ is Hausdorff, implying that $X_n$ is also Hausdorff, so that points are closed in $X_n$. Therefore, the subsequences $\{x_m,x_{m+1},\ldots\}$, $m=1,2,3$ form a nested system of closed subsets of $K$whose intersection is empty, although the intersection of every finite subsystem is nonempty. And this would give us a contradiction to the compactness of $K$.

Comments.

1. $X_n$ is Hausdorff for each $n$. Hence every finite subset of $X_n$ is closed (which is true more generally for every $T_1$ space). In particular, for every $m$ and every $n$, if $T_m=\{x_m,x_{m+1},\ldots\}$ (the set formed by taking the points in a tail of the original sequence) then $T_m\cap X_n$ is closed in $X_n$ because it is finite, because $T_m\cap X_n\subseteq\{x_m,x_{m+1},\ldots,x_{n-1}\}$ which is clearly finite. (If $n\le m$ then $\{x_m,x_{m+1},\ldots,x_{n-1}\}$ is empty, but at any rate, even when $n>m$ we have that $\{x_m,x_{m+1},\ldots,x_{n-1}\}$ is finite). If you fix $m$ we obtain that $T_m$ is closed (in $X$), because its intersection with each $X_n$ is closed (in $X_n$). Note that each $T_m$ is a subset of $K$ (since each $x_n$ is in $K$), and that $T_m$ is a closed subset of $K$, because $T_m$ is closed $X$.
2. The sequence $\{T_m:m=1,2,3,\ldots\}$ is a nested decreasing sequence of non-empty closed sets in the compact space $K$. By the finite intersection property for compact spaces, we must have that $\bigcap_{m=1}^\infty T_m$ is non-empty. But it is obviously empty, a contradiction, which completes the proof.