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Here are two proofs that compact sets in Hausdorff spaces must be closed.
1. Indirect proof:
Consider a compact set, and a boundary point of that set. Because the space is Hausdorff, there exists for every point other than that boundary point disjoint neighborhoods. If this point is not in the set, then those neighborhoods disjoint from the neighborhoods of the boundary point form a cover, and thus has a finite subcover. Consider the neighborhoods of the boundary point that are disjoint from that finite subcover. The intersection of them is an open set that does not meet the set, which contradicts the fact that the point is a boundary point of the set.
2. Direct proof
Consider a boundary point of the set. Because the space is Hausdorff, there exists for every point other than that boundary point disjoint neighborhoods. Any finite number of those neighborhoods disjoint from the neighborhoods of the boundary point does not cover the whole set because it does not cover the intersection of the the neighborhoods of the boundary point to which they correspond, which itself is a neighborhood of the neighborhood, and thus must contain a point of the set. Therefore, because the set is compact, the union of all such neighborhoods that were disjoint from the neighborhood of the boundary point cannot cover the whole set, even though it contains all elements of the set if it is not the boundary point. Therefore the boundary point must be within the set.
Here is another example:
http://www.dpmms.cam.ac.uk/~wtg10/FTA.html
It is generally desirable that once an indirect proof is found, that a direct proof be formed, which is always possible.