Let $T$ be a continuous (thus bounded) linear map from a Banach space $X$ to a Banach space $Y$. I want to prove that if $T$ has a continuous inverse, then $R_T$ is closed, where $R_T=T(X)$ is the range of $T$.
Attempt: We recall that $R_T$ is closed if its complement is open or equivalently $R_T$ is closed if it contains all its limit points. Further we recall that a function with inverse is both injective and surjective. I thought, perhaps naively, that since the function $T$ is a bijection we have that $R_T$ is all of $Y$ and hence the complement of $R_T$ is the empty set which is always open and closed in any topology. But then we don't use that the inverse of $T$ is continuous so this reasoning is probably wrong.
I also considered if we may use the open mapping theorem which say that because $T^{-1}$ is a bounded (since continuous) surjective map between Banach spaces, it maps open sets in $Y$ surjectively to open sets in $X$. But I couldn't come up an argument.
How to solve the problem?