I have a "proof" for the following problem:
Suppose $X$ is a Banach space, the operator $T \in L(X,X)$ is open, and let $X_0$ be a closed subspace of $X$. Further the restriction $T_0$ of $T$ to $X_0$ is continuous. Is $T_0$ necessarily open?
I found a counterexample here.
What went wrong in my "proof" then? --
$T_0$ is continuous $\implies$ $X_0 \times T(X_0)$ is closed by closed graph theorem $\implies T(X_0)$ is closed because graph is closed so it must be closed in each of its components $\implies T_0$ is open by open mapping (because range is closed)