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If $(X,T)$ is a Hausdorff space such that every proper closed subspace is compact, prove that $(X,T)$ is compact.

I know I have to show that $X$ has a finite subcover, but I'm not sure how to do this. I thought to let $X = \bigcup_{A \subset X} A$ such that $A$ is a closed compact subset, but I don't think that an infinite union of compact sets will always be compact.

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Suppose $x,y$ are distinct points of $X$ (if they don't exist we're done anyway..). Let $U$ and $V$ be disjoint neighbourhoods of $x$ resp. $y$.

Then $X = (X \setminus U) \cup (X \setminus V)$ (why) and both of these are proper (why?) closed subsets. Now use that a union of compact subsets is compact.

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Hint: Given an open cover ${\mathscr T}$ of $X$ and $O\in{\mathscr T}\setminus\{\emptyset\}$, consider the induced cover of $X\setminus O$ by ${\mathscr T}\setminus\{O\}$.

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A cover of a set $X$ is a family of subsets $\mathcal{F}=\{A_i : i\in I\}$ such that $X=\cup_{i\in I} A_i$. A cover of a topological space is open if every set in it is open. A topological space $X$ is compact if every open cover has a finite subcover.

Let $X$ be a top. space such that every proper closed subspace is compact and $\mathcal{F}$ be an open cover of $X$. We may assume $X$ is nonempty. If $X\in \mathcal{F}$ then $\{X\}\subseteq \mathcal{F}$ is a subcover. Suppose this is not the case, then there must be $A_0\in\mathcal{F}$ open nonempty proper subset of $X$. $B=X\setminus A_0$ is a closed proper subset of $X$ so by hypothesis it is compact and $\mathcal{F}_B=\{A \cap B : A\in \mathcal{F}\setminus \{A_0\}\} $ is an open cover of $B$ so there exists a finite subcover $\{A_1 \cap B, \ldots, A_n \cap B\}$. Now show that $\{A_0, A_1,\ldots, A_n\}$ is a cover of $X$. Hence we conclude $X$ is compact.

I do not see why we need the fact that $X$ is Hausdorff here. Also arbitrary union of compact sets need not be compact, for example take into account that a point is compact in any topological space.

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