Let $(E,\|\cdot\|)$ be a normed space. Consider the dual space $E^*$ and equip it with the weak star topology. Set $V:=(E^*,\text{wk}^*)^*=\{\phi:E^*\to\mathbb{C}: \phi\text{ is wk}^*\text{ continuous}\}$. It is obvious that, if $e\in E$, then $\phi_e:E^*\to\mathbb{C}$, $\phi_e(f):=f(e)$ is weak-star continuous, so $\{\phi_e:e\in E\}\subset V$. We will show that $V=\{\phi_e:e\in E\}$.
Now let $\phi\in V$. Note that the weak-star topology on $E^*$ is a locally convex topology generated by the seminorms $\{p_e\}_{e\in E}$, where $p_e(f):=|f(e)|$. By elementary theory of locally convex spaces (see for example the appendix in Murphy's book, theorem A.1), there exist $e_1,\dots,e_n\in E$ and $M>0$ such that
$$|\phi(f)|\le M\cdot\max_{1\le j\le n}|p_{e_j}(f)|=M\cdot\max_{1\le j\le n}|f(e_j)|$$
Therefore, if $f(e_j)=0$ for all $j=1,\dots,n$, then $\phi(f)=0$. In other words,
$$\bigcap_{j=1}^n\ker(\phi_{e_j})\subset\ker(\phi)$$
By this exercise of linear algebra this shows that $\phi$ is a linear combination of the $\phi_{e_j}$, so find $\lambda_1,\dots,\lambda_n\in\mathbb{C}$ such that $\phi=\sum_{j=1}^n\lambda_j\phi_{e_j}$. But if $e:=\sum_{j=1}^n\lambda_je_j\in E$, then $\phi_e=\sum_{j=1}^n\lambda_j\phi_{e_j}$, so $\phi=\phi_e$ as we wanted.
Finally, $V$ has its own weak-star topology, i.e. $\phi_i\to\phi$ weak-star in $V$ if and only if $\phi_i(f)\to\phi(f)$ for all $f\in E^*$. But writing $\phi_i=\phi_{e_i}$ and $\phi=\phi_e$ for $(e_i)\subset E, e\in E$, we see that $\phi_{e_i}\to\phi_e$ weak-star in $V$ if and only if $f(e_i)\to f(e)$ for all $f\in E^*$, i.e. if and only if $e_i\to e$ weakly in $E$. In other words, what we have almost just shown is that the map $(E,\text{weak})\to (V,\text{weak*})$, $e\mapsto \phi_e$ is an isomorphism of locally convex topological spaces, i.e. a linear homeomorphism. All we haven't showed yet is injectivity of this map. But indeed, if $\phi_e=0$, then $f(e)=0$ for all $f\in E^*$, and it is well-known by the Hahn-Banach theorem that this implies that $e=0$.
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
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.