Since $\phi$ is an isometry, $\phi(B_E) \subseteq B_{E^{**}}$. As $B_{E^{**}}$ is $\sigma(E^{**}, E^*)$-compact by the [Banach-Alaoglu Theorem](/theorems/212) and therefore $\sigma(E^{**}, E^*)$-closed, $\overline{\phi(B_E)}^{w^*} \subseteq B_{E^{**}}$. For the reverse inclusion, let $\psi \in B_{E^{**}}$ and let $V$ be a $\sigma(E^{**}, E^*)$-neighbourhood of $\psi$. Then $V$ contains a basic neighbourhood of the form $V = \{\eta \in E^{**} : |\eta(F_i) - \psi(F_i)| < \varepsilon, \, i = 1, \ldots, n\}$ for some $F_1, \ldots, F_n \in E^*$ and $\varepsilon > 0$. We show $V \cap \phi(B_E) \ne \varnothing$. Set $\alpha_i := \psi(F_i)$. For any $\beta_1, \ldots, \beta_n \in \mathbb{R}$: \begin{align*} \left|\sum_i \beta_i \alpha_i\right| = \left|\psi\left(\sum_i \beta_i F_i\right)\right| \le \|\psi\|_{E^{**}} \cdot \left\|\sum_i \beta_i F_i\right\|_{E^*} \le \left\|\sum_i \beta_i F_i\right\|_{E^*} \end{align*} since $\|\psi\|_{E^{**}} \le 1$. If no $f \in B_E$ satisfied $|F_i(f) - \alpha_i| < \varepsilon$ for all $i$, then $\alpha := (\alpha_1, \ldots, \alpha_n)$ would lie outside the image of the [linear map](/page/Linear%20Map) $H: E \to \mathbb{R}^n$ sending $f \mapsto (F_1(f), \ldots, F_n(f))$ restricted to $B_E$. Since $\{\alpha\}$ and $H(B_E)$ are convex and disjoint in $\mathbb{R}^n$, separation by a hyperplane gives $\beta \in \mathbb{R}^n$ and $\gamma \in \mathbb{R}$ with $\sum_i \beta_i F_i(f) < \gamma < \sum_i \beta_i \alpha_i$ for all $f \in B_E$. This gives $\|\sum_i \beta_i F_i\|_{E^*} \le \gamma < |\sum_i \beta_i \alpha_i|$, contradicting the inequality above. Hence $V \cap \phi(B_E) \ne \varnothing$, so $\psi \in \overline{\phi(B_E)}^{w^*}$.