For each , use the supplied Hausdorff condition fact to choose disjoint open sets with and . Compactness of gives finitely many whose cover . Set
Both are open; contains and contains . They are disjoint because membership in would put a point in some . If one compact set is empty, use the empty set and directly. Thus disjoint compact subspaces of a Hausdorff space admit disjoint open neighborhoods.

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