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Let $(X, d)$ be a compact metric space. Suppose that $(K_\alpha)_{\alpha \in I}$ is a collection of closed sets in $X$ with the property that any finite subcollection of these sets necessarily has non-empty intersection, thus $\bigcap_{\alpha \in F} K_\alpha \not= \emptyset$ for all finite $F \subseteq I$. (This property is known as the finite intersection property.) Show that the entire collection has non-empty intersection, thus $\bigcap_{\alpha \in I}K_\alpha \not= \emptyset$. Show by counterexample that this statement fails if $X$ is not compact.

Suppose that $\bigcap_{\alpha \in I}K_\alpha = \emptyset$. This implies that $\bigcup_{\alpha \in I}K_\alpha^c = X$. $X$ is compact and $\bigcup_{\alpha \in I}K_\alpha^c$ is an open cover of $X$. Thus, there exist a finite subcover. This implies that $I$ must be finite. But, this is contradiction due to the finite intersection property.

I am struggling to come up with a counter example. I appreciate if you give some hint.

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Let $X = (0,1)$ and $K_n = (0,\frac{1}{n}]$. Can you finish?

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