Proof

By hypothesis, $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ are topological spaces and $f:X\to Y$ is a bijective function.

Assume, $f$ is a homeomorphism, i.e. $f$ and its inverse function $f^{-1}$ are both continuous.
It suffices to show that $f$ is an open function.
By bijective open functions, $f$ is open since $f^{-1}$ is continuous.

Assume, $f$ is both, continuous and an open function.
Let $A\subseteq X$ be a subset of $X.$
Case $A$ is closed in $X.$
- Then $A=A^-$ (equals its closure in $X$).
- Since $f$ is an open function, the image $f[A^-]$ is closed in $Y.$
- Thus, $f[A^-]=f[A^-]^-$ (equals its closure in $Y$).
- Therefore, $f[A^-]=f[A]^-.$
Case $A$ is open in $X.$
- Since $f$ is continuous, by equivalent notions of continuous functions, $f[A^-]\subseteq f[A]^-.$
- Since $f$ is an open function, $f[A^-]$ is closed in $Y.$
- Therefore, there is no $y\in f[A]^-$ such that $y\not\in f[A^-].$
- In other words, $f[A^-]\subseteq f[A]^-$ and $f[A^-]\not\subset f[A]^-.$
- It follows, $f[A^-]= f[A]^-.$

Assume, for all subsets $A\subseteq X$, the image of the closure equals the closure of the image, formally $f[A^-]=f[A]^-.$
Since $f[A^-]\subseteq f[A]^-,$ the function $f$ is continuous by equivalent notions of continuous functions.
On the other hand, since $(2)$ is sufficient for the assumption $(3),$ $f$ must be an open function.
Since $f$ is an open function and bijective, its inverse function $f^{-1}$ is continuous, applying bijective open functions.
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