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.
Let $F$ be a countable dense subset of $E'$. By Hahn-Banach theorem, we need to find a countable subset $D$ of $E$ such that $$\forall f \in E' \big [ f \restriction \operatorname{span}_\mathbb R D \equiv 0 \implies f \equiv 0 \big ].$$
Clearly, $\operatorname{span}_\mathbb R D$ is a vector subspace of $E$. If $D$ satisfies above condition, then $\overline{\operatorname{span}_\mathbb R D} = E$. Clearly, $\overline{\operatorname{span}_\mathbb Q D} = \operatorname{span}_\mathbb R D$ and thus $\overline{\operatorname{span}_\mathbb Q D} = E$. Moreover, $\overline{\operatorname{span}_\mathbb Q D}$ is also countable.
Let's characterize $D$. Fix $f\in E'$ such that $f \restriction \operatorname{span}_\mathbb R D \equiv 0$. There is a sequence $(f_n)$ in $F$ such that $f_n \to f$. We have
$$
\sup_{x \text{ s.t. } |x|=1} \langle f_n-f, x \rangle = \|f_n-f\| \to 0.
$$
The hypothesis $f \restriction \operatorname{span}_\mathbb R D \equiv 0$ implies $f \restriction D \equiv 0$. It follows that
$$
\sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle \to 0.
$$
It suffices that $f_n \to 0$ or equivalently $\|f_n\| \to 0$. This suggests us a "bound"
$$
\|f_n\| \le\sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle.
$$
This "bound" maybe too strong to be true, but a weaker bound (by a positive constant $\alpha <1$) is enough for our goal, i.e.,
$$
\alpha\|f_n\| \le\sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle.
$$
For this inequality to hold, it suffices that for each $n$, there is $x_n \in D$ such that $|x_n|=1$ and
$$
\alpha\|f_n\| \le \langle f_n, x_n \rangle.
$$
Such pick of $x_n$ is possible due to $\alpha \in (0, 1)$. This completes the proof.
Best Answer
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.
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.