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.
A more direct alternative: let $\mathcal{U}$ be a subbasic cover of $E$, so that each $U \in \mathcal{U}$ is of the form $\pi_{i(U)}^{-1}[O_U]$ for some $i(U) \in I$ and some $O_U \in \mathcal{T}_{E_i}$.
If for some $i \in I$ we have that $\{O_U\mid i(U)=i\}$ is an open cover of $E_i$, by compactness of $E_i$ we can find a finite subcover of $\mathcal{U}$.
So we can assume WLOG that this is not the case and using AC we can pick $p_i \in E_i\setminus \bigcup \{O_U\mid i(U)=i\}$. This $p$ is then covered by no member of $\mathcal{U}$. Contradiction, so we must actually be in the previous case and we're done.
Best Answer
By axiom of choice, $\mathfrak a$ can be well-ordered. WLOG, we assume $\mathfrak a$ is an ordinal number. Let $(f_d)_{d\in D}$ be a net in $E$ with $f_d := (f_d^\lambda)_{\lambda < \mathfrak a}$. By transfinite recursion, we will define a sequence $(T_\lambda)_{\lambda \le \mathfrak a}$ of monotonic cofinal maps $T_{\lambda}: D_{\lambda} \to D$ between directed sets such that $(f_{T_\lambda(d)}^\eta)_{d\in D_\lambda}$ is convergent for all $\eta < \lambda$.
We initialize with $D_0 : = D$ and $T_0: d \mapsto d$. Let $\mathfrak b \le \mathfrak a$. Assume that $T_\lambda$ is already defined for all $\lambda < \mathfrak b$. Our goal is to define $T_{\mathfrak b}$.
Let $\mathfrak b$ be a successor ordinal. Then there is an ordinal $\mathfrak c$ such that $\mathfrak c+1 = \mathfrak b$. By inductive hypothesis, $T_{\mathfrak c} :D_{\mathfrak c} \to D$ is already defined. Let $h_d := f_{T_\mathfrak c(d)}^\mathfrak b$. Clearly, $(h_d)_{d\in D_\mathfrak c}$ is a net in $E_\mathfrak b$. Because $E_\mathfrak b$ is compact, there is a monotonic cofinal map $K:D_{\mathfrak b} \to D_{\mathfrak c}$ such that $(h_{K(d)})_{d \in D_\mathfrak b}$ is convergent. Let $T_\mathfrak b := K \circ T_\mathfrak c$.
Let $\mathfrak b$ be a limit ordinal. Let $D_{\mathfrak b} := \{(d, \lambda) \mid d \in D_\lambda, \lambda <\mathfrak b\}$. We endow $D_{\mathfrak b}$ with a partial order $\le$ defined by $$(d_1, {\lambda_1}) \le (d_2, {\lambda_2}) \iff T_{\lambda_1} (d_1) \le T_{\lambda_1} (d_2) \text{ and } \lambda_1 < \lambda_2.$$
Let $(d_1, {\lambda_1}), (d_2, {\lambda_2}) \in D_{\mathfrak b}$. WLOG, we assume $\lambda_1 < \lambda_2$. Because $\mathfrak b$ is a limit ordinal, there is $\lambda_3 < \mathfrak b$ such that $\lambda_2 < \lambda_3$. Because $D$ is a directed set, there is $d^\star \in D$ such that $T_{\lambda_1} (d_1) \le d^\star$ and $T_{\lambda_2} (d_2) \le d^\star$. Because $T_{\lambda_3}$ is cofinal, there is $d_3 \in D_{\lambda_3}$ such that $d^\star \le T_{\lambda_3} (d_3)$. Hence $(d_1, {\lambda_1}) \le (d_3, {\lambda_3})$ and $(d_2, {\lambda_2}) \le (d_3, {\lambda_3})$. It follows that $(D_{\mathfrak b}, \le )$ is a directed set.
Let $T_\mathfrak b: D_{\mathfrak b} \to D, (d, \lambda) \mapsto T_\lambda (d)$. Clearly, $T_\mathfrak b$ is monotonic. For $d^\star \in D$, there is $d \in D_{0}$ such that $d^\star \le T_0(d)$ because $T_0$ is cofinal. Then $d^\star \le T_\mathfrak b (d, 0)$ and thus $T_\mathfrak b$ is cofinal. This completes the construction of such required transfinite sequence. Finally, we just pick up $(f_{T_\mathfrak a(d)})_{d\in D_{\mathfrak a}}$.
Update: I have just figured out that the construction in the case of limit ordinal does not guarantee the convergence of coordinates.