Proof that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism

Proof

We show that the inverse map is continuous by testing closed sets.

1. Since $f$ is bijective, it has an inverse function $g = f^{-1} \colon Y \to X$. By the definition of a homeomorphism, it remains to show that $g$ is continuous.
2. By continuity via closed sets, it is enough to show that $g^{-1}(F)$ is closed in $Y$ for each closed $F \subseteq X$.
3. Let $F \subseteq X$ be closed. The preimage is $g^{-1}(F) = \set{y \in Y : g(y) \in F}$.
4. For $y \in Y$, we have $g(y) \in F$ if and only if $y = f(x)$ for some $x \in F$, because $g(y)$ is the unique $x$ with $f(x) = y$. So $g^{-1}(F) = f(F)$.
5. The subspace topology on $X$ as a subset of itself is the topology of $X$. So $X$ is a compact set in $X$.
6. Since closed subsets of compact sets are compact, $F$ is compact.
7. The continuous image of a compact set is compact, so $f(F)$ is compact in $Y$.
8. Since $Y$ is Hausdorff, compact subsets of a Hausdorff space are closed. So $f(F)$ is closed in $Y$.
9. By steps 4 and 8, $g^{-1}(F)$ is closed for each closed $F$. So $g$ is continuous, and $f$ is a homeomorphism.