I want to prove that given $X$ reflexive if $Y=X\oplus_2 X'$ is Hilbert then $X$ is Hilbert. I showed already that if $X$ is Hilbert then $X'$ (the dual) is also Hilbert by Frechet-Reisz representation theorem. To conclude I want to suppose by contradiction that $X$ is not Hilbert. Then $X'$ is not Hilbert as if $X'$ was Hilbert then $X''\cong X$ would be Hilbert which is a contradiction. So now we have that $Y=X\oplus_2 X'$ is the direct sum of two non Hilbert spaces. So I would like to say that $Y$ cannot be Hilbert as well.
Is there a known result that states this? Otherwise how could I prove this statement?
ps: I know that an easier way would be to show that $X$ inherits the parallelogram law by $||x||^2=||(x,0)||^2$ but I wanted to ask about the general result through this question.
Thanks in advance