Suppose that $X$ and $Y$ are Banach spaces, and $D ⊂ X$ is a linear subspace, which may not be closed. Suppose that $T : D → Y$ has a closed graph (in $X\times Y$), and is $1-1$ and onto. If $D$ is not closed, then $T$ need not be continuous. Prove, however, that $T^{−1} : Y → X$ is continuous.
This should be true for any linear spaces $D$ since you can always take $X$ to be the completion of $D$. I just don't know how to use the property of closed graph, since $D$ is not necessarily Banach..
In fact I misunderstood the problem at a first glance. I thought T has a closed graph in $D\times Y$ but not necessarily in $X\times Y$.. Now it's just a direct application of closed graph theorem.
Note that $T$ is equal to the composite $D\rightarrow\Gamma(T)\rightarrow Y$ (with $\Gamma(T)\rightarrow Y$ the projection, which is bijective and continuous), and that since $T$ has closed graph $\Gamma(T)$, $\Gamma(T)$ is a Banach space (for it is a closed subspace of the Banach space $X\times Y$). Then $T^{-1}$ is the composite of continuous maps $Y\rightarrow\Gamma(T)\rightarrow D$ with $Y\rightarrow\Gamma(T)$ the inverse of the projection (continuous by general principles).