I have a theorem telling me that some property holds for operators that are bounded and weakly sequentially closed. Somehow, I have in mind that boundedness actually implies the weakly sequentially closedness. My reasoning would be this: As boundedness implies closedness and since convex sets (graph is convex) are weakly sequentially closed if they are closed, the second property should be redundant.
A bounded operator is closed if and only if its domain is, so your implication "boundedness implies closedness" is not true in general. See here for more details.