Range of operator always closed. Mistake in argument

Question

Let $A \in L(X,Y)$ be a linear operator between Hilbert spaces and the operator $$\hat{A}: \ker(A)^{\perp} \rightarrow \operatorname{ran}(A)$$ is a restriction of $A$ which is bijective. Now $\ker(A)^{\perp}$ is certainly a closed space and since this operator is bijective and continuous, I would also suspect that this means that $\operatorname{ran}(A)$ is closed, because $\hat{A}$ is an isomorphism. But of course not every range of an operator is closed, so there has to be a mistake in this argument.

Edit: Ah got it. The inverse operator does not need to be bounded, as the open mapping theorem only works if $\operatorname{ran}(A)$ is closed, too.

Answer 1

The inverse operator does not need to be continuous, so we cannot conclude that the range is closed.

In fact, if the range is closed, this immediately implies that the inverse operator is continuous by the open mapping theorem.