I'm reading the proof of Kakutani's theorem in Brezis's book of Functional Analysis.
Theorem 3.17 (Kakutani). Let $E$ be a Banach space. Then $E$ is reflexive if and only if $$B_{E}=\{x \in E ;\|x\| \leq 1\}$$ is compact in the weak topology $\sigma\left(E, E^{\star}\right)$.
The direction "$\rightarrow$" relies on Banach–Alaoglu theorem, while the reverse "$\leftarrow$" relies on Helly's and Goldstine's theorems. All three theorems (Kakutani's, Helly's, and Goldstine's) are stated for Banach spaces. However, I could not see how completeness is used in the proofs. I think that they hold for general normed linear spaces.
I posted the full proofs from the textbook below. Could you confirm if my understanding is correct?
Proof of Kakutani's theorem. Assume first that $E$ is reflexive, so that $J\left(B_{E}\right)=B_{E^{\star \star}}$. We already know (by Banach-Alaoglu theorem) that $B_{E^{\star \star}}$ is compact in the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$. Therefore, it suffices to check that $J^{-1}$ is continuous from $E^{\star \star}$ equipped with $\sigma\left(E^{\star \star}, E^{\star}\right)$ with values in $E$ equipped with $\sigma\left(E, E^{\star}\right)$. In view of Proposition 3.2, we have only to prove that for every fixed $f \in E^{\star}$ the map $\xi \mapsto\left\langle f, J^{-1} \xi\right\rangle$ is continuous on $E^{\star \star}$ equipped with $\sigma\left(E^{\star \star}, E^{\star}\right)$. But $\left\langle f, J^{-1} \xi\right\rangle=\langle\xi, f\rangle$, and the map $\xi \mapsto\langle\xi, f\rangle$ is indeed continuous on $E^{\star \star}$ for the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$. Hence we have proved that $B_{E}$ is compact in $\sigma\left(E, E^{\star}\right)$.
The converse is more delicate and relies on the following two lemmas:
Lemma 3.3 (Helly). Let $E$ be a Banach space. Let $f_{1}, f_{2}, \ldots, f_{k}$ be given in $E^{\star}$ and let $\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k}$ be given in $\mathbb{R}$. The following properties are equivalent:
(i) $\forall \varepsilon>0 \exists x_{\varepsilon} \in E$ such that $\left\|x_{\varepsilon}\right\| \leq 1$ and $$ \left|\left\langle f_{i}, x_{\varepsilon}\right\rangle-\gamma_{i}\right|<\varepsilon \quad \forall i=1,2, \ldots, k, $$ (ii) $\left|\sum_{i=1}^{k} \beta_{i} \gamma_{i}\right| \leq\left\|\sum_{i=1}^{k} \beta_{i} f_{i}\right\| \quad \forall \beta_{1}, \beta_{2}, \ldots, \beta_{k} \in \mathbb{R}$.
Proof of Helly's theorem. (i) $\Rightarrow$ (ii). Fix $\beta_{1}, \beta_{2}, \ldots, \beta_{k}$ in $\mathbb{R}$ and let $S=\sum_{i=1}^{k}\left|\beta_{i}\right|$. It follows from (i) that $$ \left|\sum_{i=1}^{k} \beta_{i}\left\langle f_{i}, x_{\varepsilon}\right\rangle-\sum_{i=1}^{k} \beta_{i} \gamma_{i}\right| \leq \varepsilon S $$ and therefore $$ \left|\sum_{i=1}^{k} \beta_{i} \gamma_{i}\right| \leq\left\|\sum_{i=1}^{k} \beta_{i} f_{i}\right\|\left\|x_{\varepsilon}\right\|+\varepsilon S \leq\left\|\sum_{i=1}^{k} \beta_{i} f_{i}\right\|+\varepsilon S . $$ Since this holds for every $\varepsilon>0$, we obtain (ii).
(ii) $\Rightarrow$ (i). Set $\gamma=\left(\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k}\right) \in \mathbb{R}^{k}$ and consider the map $\varphi: E \rightarrow \mathbb{R}^{k}$ defined by $$ \varphi(x)=\left(\left\langle f_{1}, x\right\rangle, \ldots,\left\langle f_{k}, x\right\rangle\right) $$ Property (i) says precisely that $\gamma \in \overline{\varphi\left(B_{E}\right)}$. Suppose, by contradiction, that (i) fails, so that $\gamma \notin \overline{\varphi\left(B_{E}\right)}$. Hence $\{\gamma\}$ and $\overline{\varphi\left(B_{E}\right)}$ may be strictly separated in $\mathbb{R}^{k}$ by some hyperplane; i.e., there exists some $\beta=\left(\beta_{1}, \beta_{2}, \ldots, \beta_{k}\right) \in \mathbb{R}^{k}$ and some $\alpha \in \mathbb{R}$ such that $$ \beta \cdot \varphi(x)<\alpha<\beta \cdot \gamma \quad \forall x \in B_{E} . $$ It follows that $$ \left\langle\sum_{i=1}^{k} \beta_{i} f_{i}, x\right\rangle<\alpha<\sum_{i=1}^{k} \beta_{i} \gamma_{i} \quad \forall x \in B_{E}, $$
and therefore $$ \left\|\sum_{i=1}^{k} \beta_{i} f_{i}\right\| \leq \alpha<\sum_{i=1}^{k} \beta_{i} \gamma_{i} $$ which contradicts (ii).
Lemma 3.4 (Goldstine). Let $E$ be any Banach space. Then $J\left(B_{E}\right)$ is dense in $B_{E^{\star \star}}$ with respect to the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$, and consequently $J(E)$ is dense in $E^{\star \star}$ in the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$.
Proof of Goldstine's theorem. Let $\xi \in B_{E^{\star \star}}$ and let $V$ be a neighborhood of $\xi$ for the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$. We must prove that $V \cap J\left(B_{E}\right) \neq \emptyset$. As usual, we may assume that $V$ is of the form $$ V=\left\{\eta \in E^{\star \star} ;\left|\left\langle\eta-\xi, f_{i}\right\rangle\right|<\varepsilon \quad \forall i=1,2, \ldots, k\right\} $$ for some given elements $f_{1}, f_{2}, \ldots, f_{k}$ in $E^{\star}$ and some $\varepsilon>0$. We have to find some $x \in B_{E}$ such that $J(x) \in V$, i.e., $$ \left|\left\langle f_{i}, x\right\rangle-\left\langle\xi, f_{i}\right\rangle\right|<\varepsilon \quad \forall i=1,2, \ldots, k . $$ Set $\gamma_{i}=\left\langle\xi, f_{i}\right\rangle$. In view of Lemma $3.3$ it suffices to check that $$ \left|\sum_{i=1}^{k} \beta_{i} \gamma_{i}\right| \leq\left\|\sum_{i=1}^{k} \beta_{i} f_{i}\right\|, $$ which is clear since $\sum_{i=1}^{k} \beta_{i} \gamma_{i}=\left\langle\xi, \sum_{i=1}^{k} \beta_{i} f_{i}\right\rangle$ and $\|\xi\| \leq 1$.
Proof of Theorem 3.17, concluded. The canonical injection $J: E \rightarrow E^{\star \star}$ is always continuous from $\sigma\left(E, E^{\star}\right)$ into $\sigma\left(E^{\star \star}, E^{\star}\right)$, since for every fixed $f \in E^{\star}$ the map $x \mapsto\langle J x, f\rangle=\langle f, x\rangle$ is continuous with respect to $\sigma\left(E, E^{\star}\right)$. Assuming that $B_{E}$ is compact in the topology $\sigma\left(E, E^{\star}\right)$, we deduce that $J\left(B_{E}\right)$ is compact-and thus closed一in $E^{\star \star}$ with respect to the topology $\sigma\left(E^{\star \star}, E^{\star}\right)$. On the other hand, by Lemma 3.4, $J\left(B_{E}\right)$ is dense in $B_{E^{\star \star}}$ for the same topology. It follows that $J\left(B_{E}\right)=$ $B_{E^{\star \star}}$ and thus $J(E)=E^{\star \star}$.