Continuous bijection from a compact space to a Hausdorff space is a homeomorphism

Statement

Let $X$ be a compact space, let $Y$ be a Hausdorff space and let $f \colon X \to Y$ be a continuous map that is bijective. Then $f$ is a homeomorphism.

Remarks

Proof: it is enough to show that $f$ maps closed sets to closed sets. A closed $F \subseteq X$ is compact because closed subsets of compact sets are compact. Then $f(F)$ is compact, and thus closed in $Y$ by compact subsets of a Hausdorff space are closed. The compactness is needed: the map $t \mapsto (\cos t, \sin t)$ from $[0, 2\pi)$ to the unit circle is a continuous bijection, but its inverse is not continuous.

Explanations

L1 human

For compact spaces, you do not need to check that the inverse map is continuous: it comes for free. This is a quick way to show that two compact spaces have the same shape.