Proof. ($F$): Assume that $E$ is reflexive. As $F$ is closed, it is weakly closed. Thus $\ol{B_F(0, 1)}= \ol{B_E(0, 1)}\cap F$ is weakly compact. By (2) of Theorem 14.1.10, $E$ is reflexive.

($E/F$): Assume that $E$ is reflexive. Let $\pi: E \to E/F$ be the canonical map, then $\pi$ is also weakly continuous. Hence $\pi(\ol{B_E(0, 1)})$ is weakly compact in $E/F$. By (2) of Theorem 14.1.10, $E/F$ is reflexive.

($E$): Assume that $F$ and $E/F$ are reflexive. Let $x_{0} \in E^{**}$, and $T: F^{\perp} \to (E/F)^{*}$ be the canonical isomorphism, then $x_{0} \circ T^{-1}\in (E/F)^{**}$. Given that $E/F$ is reflexive, there exists $x \in E$ such that $x_{0} \circ T^{-1}= x + F$. In particular, for any $\phi \in F^{\perp}$,

so $x_{0} - x \in F^{\perp\perp}$. By the Bipolar Theorem, $F^{\perp\perp}= F^{**}= F$. Therefore $x_{0} - x \in F$, and $x_{0} \in E$.$\square$