Let $(X,|.|)$ be a Banach space. We know that if $A:X\to X$ is a closed operator then $(X,|.|_A)$ is a Banach space, where $|.|_A$ is the norm defined by $$|x|_A=|x|+|Ax|$$ Then using the "continuity of the inverse theorem", we can show that $A$ is even a bounded operator. This is actually the proof of the closed graph theorem.
Now if we assume that $(X,|.|_A)$ is a Banach space, is there a simple proof to show that $A$ is closed without using the "continuity of the inverse theorem" ?
$x \mapsto (x,Ax)$ is an isometry between $(X, \lvert\,\cdot\lvert_A)$ and the graph
$$\Gamma(A) = \{ (x,Ax) : x \in X\} \subset X\times X$$
of $A$ when we endow $X\times X$ with the norm $\lVert (x,y)\rVert = \lvert x\rvert + \lvert y\rvert$. That $(X,\lvert\,\cdot\rvert_A)$ is a Banach space thus means $\Gamma(A)$ is a complete subspace of $X\times X$. But a complete subspace of a Hausdorff space is always closed, hence it means $\Gamma(A)$ is a closed subspace of $X\times X$, which by definition means $A$ is closed.