# Let $E$ be a reflexive Banach space and $M$ its closed linear subspace. Then $M$ is reflexive

I'm trying to re-phrase this proof in Brezis's book of Functional Analysis.

Let $$E$$ be a reflexive Banach space and $$M$$ its closed linear subspace. Then $$M$$ is reflexive.

Could you have a check if I understand and apply concept of compactness/subspace topology correctly?

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, then I will happily accept theirs.

By Kakutani's theorem, $$M$$ is reflexive if and only if $$B_M := \{x\in M | |x| \le 1\}$$ is compact in the weak topology $$\sigma(M, M')$$. Notice that $$\sigma(M, M')$$ coincides with the subspace topology that $$\sigma(E, E')$$ induces on $$M$$. So $$M$$ is reflexive if and only if $$B_M$$ is compact in $$\sigma(E, E')$$.
Also by Kakutani's theorem, we have $$B_E$$ is compact in $$\sigma(E,E')$$. A closed subset of a compact set is also compact, so it suffices to show that $$B_M$$ is closed in $$B_E$$ w.r.t. $$\sigma(E, E')$$. This is indeed true because $$B_M = M \cap B_E$$ where $$M$$ is convex and closed in norm topology, and thus closed in $$\sigma(E, E')$$.