On the closedness of KdV operator

Question

Some papers that I am currently reading state that the classical KdV operator $Au=u'+u'''$ with $D(A)=\{u\in H^3(0,1)|u(0)=u(1)=u'(1)=0\}$ is a closed operator in $L^2(0,1)?$ However, no proof is given. How do I show this fact?

Answer 1

I came up with an answer which uses the sequential definition of closedness. It can be done via a compactness argument but thanks anyway for the effort.