$f$ is a homeomorphism iff $f$ is bijective, continuous and open

Continuing where you stopped: Since $f$ is bijective, we have that $f(f^{-1}(U)) = U$ and since $U$ is open, we have that $f(V)$ is open (in $Y$)

To prove that $f^{-1}$ is continuous, we should prove that for all open sets $U \in X$, $(f^{-1})^{-1}(U)$ is open in $Y$, since $(f^{-1})^{-1}(U) = f(U)$. And since $f$ is open, this follows directly.