I. This is true. The continuous image of a compact set is compact. Take any open cover
of Y. Then
so
is an open cover of X. Since X is compact, you can take a finite subcover
. Then
covers Y. Thus any open cover of Y has a finite subcover.
II. This is false. Take the set {1,2,3}. Give it the discrete topology on X, and give it some non-Hausdorff topology on Y, e.g. with open sets {1,2,3}, {}, {1}, {2,3}. Then
defined by f(1)=1, f(2)=2, f(3)=3 is a continuous bijection, since by the discrete topology every subset of X is open so
is open for all U open in Y.
III. This is true. To prove that the inverse of f is continuous, we can show that the inverse of the inverse of f maps closed sets to closed sets. So we need to show that f(F) is closed for every closed subset F in X. Since F is a closed subset in a compact space X, then F is compact. Therefore by continuity, f(F) is compact, since the image of a compact set under a continuous map is compact. Since compact subsets of a Hausdorff space are closed, we have that f(F) is closed.