Because $\varphi$ is continuous at $0$ in $\sigma(E^\star, E)$, there is a neighborhood $U$ of $0$ in $\sigma(E^\star, E)$ such that $$| \langle \varphi, f \rangle | < 1, \quad \forall f\in U.$$
By construction of $\sigma(E^\star, E)$, there are $x_1, \ldots, x_n \in E$ such that $$U \supseteq V := \{f \in E^\star \mid \forall k = 1, \ldots, n: |\langle Jx_k, f \rangle| < 1\}.$$
Let $V_k := V/k := \{f/k \mid f \in V\}$ and $U_k := U/k$. It's easy to show that $$\bigcap_{k=1}^n \ker (Jx_k) = \bigcap_{k \in \mathbb N^\star} V_k \subseteq \bigcap_{k \in \mathbb N^\star} U_k = \ker \varphi.$$
It follows that $\varphi$ is linearly dependent of $Jx_1, \ldots, Jx_n$. This means $\varphi = \lambda_1 Jx_1 + \cdots + \lambda_n Jx_n = J (\lambda_1 x_1 + \cdots + \lambda_n x_n)$. Hence $e:= \lambda_1 x_1 + \cdots + \lambda_n x_n$ satisfies the requirement.
Let $F := \mathbb R^E$ be the collection of all maps from $E$ to $\mathbb R$. Let $\mathbb R_x := \mathbb R$ for all $x\in E$. Then we can write $$F = \prod_{x\in E} \mathbb R_x.$$ In this way, we endow $F$ with the product topology. Of course, $E^\star \subseteq F$. Let $i:E^\star_\mathrm{w} \to F, f \mapsto f$ be the canonical injection. Let's prove that $i$ is continuous. For $x\in E$, let $\pi_x: F \to \mathbb R_x, f \mapsto f(x)$ be the canonical projection. It suffices to show that $\pi_x \circ i:E^\star_\mathrm{w} \to \mathbb R_x, f \mapsto \langle f, x \rangle$ is continuous for all $x\in E$. This is clearly true due to the construction of the weak$^\star$ topology.
Lemma: Let $(X, \tau)$ be a topological space, $A \subseteq X$, and $\tau_A$ the subspace topology of $A$. Let $a\in A$ and $(x_d)_{d \in D}$ is a net in $A$. Then $x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$.
Clearly, $\operatorname{im} i = E^\star$. We denote by $E^\star_\tau$ the set $E^\star$ together with the subspace topology $\tau$ induced from $F$. Then $i:E^\star_\mathrm{w} \to E^\star_\tau$ is bijective. Let $f\in E^\star_\mathrm{w}$ and $(f_d)_{d\in D}$ be a net in $E^\star_\mathrm{w}$ such that $f_d \to f$. Because $i:E^\star_\mathrm{w} \to F$ is continuous, $f_d \to f$ in the topology of $F$. By our lemma, $f_d \to f$ in $E^\star_\tau$. Hence $i:E^\star_\mathrm{w} \to E^\star_\tau$ is indeed continuous.
Let $i^{-1}:E^\star_\tau \to E^\star_\mathrm{w}$ be the inverse of $i:E^\star_\mathrm{w} \to E^\star_\tau$. Let's prove that $i^{-1}$ is continuous. It suffices to show that $\varphi_x: E^\star_\tau \to \mathbb R, f \mapsto \langle f, x\rangle$ is continuous for all $x\in E$. This is indeed true because $\varphi_x = \pi_x \restriction E^\star$. Notice that continuous map sends compact set to compact set. Hence it suffices to prove that $\mathbb B_{E^\star}$ is compact in $\tau$. By our lemma, it suffices to prove that $\mathbb B_{E^\star}$ is compact in the topology of $F$.
Let $B_1 := \{f\in F \mid f \text{ is linear}\}$ and $B_2 := \prod_{x\in E}[-|x|, |x|]$. Then $\mathbb B_{E^\star} = B_1 \cap B_2$. The closed interval $[-|x|, |x|]$ is clearly compact. By Tychonoff's theorem, $B_2$ is compact.
Let $f\in F$ and $(f_d)_{d\in D}$ be a net in $B_1$ such that $f_d \to f$. Because convergence in product topology is equivalent to pointwise convergence, we get $f_d(x) \to f(x)$ for all $x\in E$. Then $f_d(x) + f_d(y) =f_d(x+y) \to f(x+y)$. On the other hand, $f_d(x) \to f(x)$ and $f_d(y) \to f(y)$. This implies $f(x+y)=f(x)+f(y)$. Similarly, $f(\lambda x) =\lambda f(x)$ for all $\lambda \in \mathbb R$. Hence $B_1$ is closed. The intersection of a closed set and a compact set is again compact. This completes the proof.
Best Answer
Let $\overline{J[B_E]}$ be the closure of $J[B_E]$ in the subspace topology that $\sigma(E'', E')$ induces on $B_{E''}$. By Banach–Alaoglu theorem, $B_{E''}$ is compact. Compact sets are closed in Hausdorff topology, so $B_{E''}$ is closed in $\sigma(E'', E')$. It follows that $\overline{J[B_E]}$ is the closure of $J[B_E]$ in $\sigma(E'', E')$.
Assume the contrary that $\overline{J[B_E]} \subsetneq B_{E''}$. Then there is $\varphi \in B_{E''} \setminus \overline{J[B_E]}$. Singletons are compact in Hausdorff topology, so $\{\varphi\}$ is compact in $\sigma(E'', E')$. Clearly, $\{\varphi\}$ and $B_{E''}$ are non-empty convex. Notice that $E''$ together with $\sigma(E'', E')$ is a locally convex t.v.s., so we can apply Hahn–Banach theorem to strictly separate $\{\varphi\}$ and $\overline{J[B_E]}$.
This means there exist $s, t \in \mathbb R$ and a linear map $\Psi:E'' \to \mathbb R$ that is continuous in $\sigma(E'', E')$ such that $$ \langle \Psi , \psi \rangle < s <t < \langle \Psi, \varphi \rangle, \quad \forall \psi \in \overline{J[B_E]}. $$
By our Lemma, there is $f \in E'$ such that $\langle \Psi, \psi \rangle = \langle \psi, f \rangle$ for all $\psi \in E''$. Then $$ \langle \psi, f \rangle < s <t< \langle \varphi , f \rangle, \quad \forall \psi \in \overline{J[B_E]}. $$
First, $t < \langle \varphi , f \rangle\le \|\varphi\| \cdot \|f\| \le\|f\|$ because $\varphi \in B_{E''}$. Second, $\langle \psi, f \rangle < s$ for all $\psi \in \overline{J[B_E]}$ implies $\langle Jx, f \rangle = \langle f, x \rangle< s$ for all $x\in B_E$. This in turn implies $\|f\| \le s$. Hence we obtain a contradiction $$\|f\| \le s<t \le \|f\|.$$
This completes the proof.