An important corollary is that a continuous 1-to-1 map of a compact space onto a hausdorff space is bicontinuous. Any closed set in the domain is also compact, its continuous image is compact, and its image is closed. Taking complements, the image of open sets is open. Thus the function is bicontinuous. Inother words, a homeomorphism. The two spaces are topologically equivalent.
The word onto is important here. Embed the closed interval [0,1] into the real line, and the result is not bicontinuous. The interval [0,½) is open in the domain, but its image is not open in the range.
Let f map X onto Y and let g map Y into Z. Let X be compact and let Y be Hausdorff. Let the functions f and g(f) be continuous. Start with a closed set in Z and take its preimage under g. If this preimage is not closed in Y, the prepreimage cannot be closed in X, since f maps closed sets to closed sets. Yet f(g) is continuous, hence both preimages are closed, and g is continuous.
Let S be hausdorff and let U be a compact subspace of S. If U is also dense in S then U = S. Remember that a compact set in a hausdorff space is closed, so U is closed. If U misses the point x then U must miss some open set containing x, which contradicts the fact that U is dense. Thus U contains every x, and U = S.
Let S be a space that is compact and hausdorff. watch what happens to S with a stronger or weaker topology. Let f map S onto itself, but let the domain have a stronger topology. Open sets still have open sets as preimage, so f is continuous. Yet some open sets in the domain are new, and don't map to open sets in S, hence f is not bicontinuous. We already showed that a map from a compact space onto a hausdorff space is a homeomorphism, and that isn't the case here; hence the domain of f is not compact. Similarly, when g maps S onto itself with a weaker topology, g is not bicontinuous, and the range is not hausdorff.
A compact hausdorff space walks a fine line. Add more open sets and lose compactness. Take away open sets and the space is no longer hausdorff.
Let T be another closed set, disjoint from S. For every x in T, separate x and S in disjoint open sets. Again, a finite cover will do. A finite intersection of open sets contains S, while a finite union of open sets contains T, and the space is normal.
Consider all possible continuous functions from S into [0,1]. These form a composite function f that maps S into the product space P, a hypercube with the weak product topology.
Select any pair of points x and y in S. Because S is normal, Urysohn's lemma provides a function fj that maps x to 0 and y to 1. Since x and y attain different values on at least one coordinate, f(x) ≠ f(y), and f is 1-1. In other words, f embeds S into P, and the map is continuous. Since the domain is compact and the range hausdorff, f is a homeomorphism.