Let $X,Y$ be topological spaces. Let $X$ be compact and let $f:X\to Y$ be continuous and bijective. Show that $f$ is a homeomorphism.
My solution: Since $f$ is continuous and bijective already the only thing left to prove is that $f^{-1}:Y\to X$ is continuous. To do that, chose an arbitrary open set $U\subset X$. Then since $f$ is continuous we know $(f^{-1}(U))^{-1}=f(U)$ must be open in $Y$.
I am not sure if this proof is legit as I havent used that fact that $X$ is compact.
As you have observed, the compactness of $X$ has to be used. Instead of open sets, Let's look at closed sets. Let $V$ be a closed set in $X$. Then $V$ is compact in $X$, as $X$ is compact. Since $f$ is continuous, $f(V)$ is compact in $Y$. Since $Y$ is hausdroff, $f(V)$ is closed in $Y$. Thus $f^{-1}$ is continuous, by the equivalent criterion of continuity using closed sets.
Hausdroff is necessary: Let $f$ be the identity function. Then $f: X \to Y$ be a bijective continuous function. If $X=[0,1]$ with the standard topology and $Y=[0,1]$ with the indiscrete topology we have that $f:X \to Y$ is not a homeomorphism.Take any open set e.g ,$U=(\dfrac{1}{4},\dfrac{1}{3})$, then $f(U)=U$ which is not open in $Y$.
