Let $X$ be a Banach space and $T: X\to \mathbb{R}$ be an unbounded linear operator. Let $S:\textrm{graph}(T)\to X, (x, Tx)\mapsto x$. Then S is bijective and continuous, but not open.
By definition of $\textrm{graph}(T)$, $x=S((x, Tx))=S((y, Ty))=y$, hence $S$ is injective. If $x\in X$ then $S((x, Tx))=x$ and therefore $S$ is surjective. Since $T$ is linear, $\textrm{graph}(T)$ is a subspace of $X\times\mathbb{R}$ (and in particular a space). $S$ is the projection to the first component and therefore continuous in the product topology. How do I see that $S$ is not open?