Using Kakutani's Theorem to prove that reflexivity is invariant under homeomorphism. I was reviewing an exercise which is
Exercise: Let $E$ and $F$ be normed spaces and suppose that $T:E \to F$ is an homeomorphism. Show that if $E$ is reflexive, then $F$ is reflexive.
I already did the exercise using the definition of reflexivity (evaluation map $J:E \to E^{\ast\ast}$ is surjective), but I was willing to rewrite using Kakutani's Theorem: 
Kakutani's Theorem: The unit closed ball is weakly compact iff the space is reflexive.
I couldn't write it down, I was wondering if you could help me. I tried something along the lines: 


*

*$T$ is continuous, so it is weakly continuous (the same for the inverse). 

*$T(\overline{B_E})$ is weakly compact.

*Try to insert $\overline{B_F}$, which is weakly closed, inside an $n\overline{B_E}$ for some $n$ and finish. 


but as you can see, I couldn't close the argument.  
EDIT: The question is answered in the comments below the accepted one. The full exercise has many claims and this was one of them. However, for this particular question the linearity is needed as @treedetective points out. Assuming this, @triple_sec's solution is correct. We concluded that there was an ambiguity in the question. Thank you for your time. 

Since $T^{-1}$ is continuous, it is also bounded: there exists some $n\in\mathbb N$ such that $\|T^{-1}(y)\|\leq n\|y\|$  for any $y\in F$.
Pick any $y\in\overline{B_F}$ and let $x\equiv T^{-1}(y)\in E$. Then, $$\|x\|=\|T^{-1}(y)\|\leq n\|y\|\leq n,$$ so that $x\in n\overline{B_E}$. Therefore, $y=T(x)\in T(n\overline{B_E})=nT(\overline{B_E})$, the last equality stemming from linearity.
It follows that $\overline{B_F}\subseteq nT(\overline{B_E})$.
