Problem 5.31 of Folland's Functional Analysis provides such an example.
Let $Y$ be a Banach space, $Z$ any normed space, and let $S : Y \to Z$ be an unbounded, everywhere defined linear operator (you need the axiom of choice to construct such a thing). Let $X \subset Y \times Z$ be the graph of $S$: $X = \{(y,Sy) : y \in Y\}$; by the closed graph theorem $X$ is not complete. Let $T : X \to Y$ be the map $T(y,Sy)=y$. $T$ is bijective and bounded but its inverse cannot be bounded, so it is not open.