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.
Let $E',E''$ be the dual and bidual of $E$ respectively. Let $B_E,B_{E''}$ be the closed unit balls of $E, E''$ respectively. Let $J:E \to E'', x \mapsto \hat x$ the canonical injection. Fix $\varphi \in B_{E''}$ and $\varepsilon>0$. WLOG, we assume $\|\varphi\| = 1$. Let $\delta$ be the modulus of uniform convexity. We pick $f\in E'$ such that $\|f\| = 1$ and $\langle \varphi, f \rangle > 1 -\delta/2$. Then
$$
U:= \{\phi \in B_{E''} \mid \langle \phi, f \rangle > 1- \delta/2 \}
$$
is an open neighborhood (nbh) of $\varphi$ in the subspace topology $\tau$ that $\sigma(E'', E')$ induces on $B_{E''}$. By Goldstine theorem, $F \cap U \neq \emptyset$ with $F:=J[B_E]$. We're done if we can prove $\operatorname{diam} U \le \varepsilon$. By Goldstine theorem again, $B_{E''} = \overline{F}^{\tau}$, so
$$
\operatorname{diam} U =\operatorname{diam} U \cap B_{E''} = \operatorname{diam} U \cap \overline{F}^{\tau} =: r_1.
$$
It follows from $U$ is $\tau$-open that $U \cap \overline{F}^{\tau} \subseteq \overline{U \cap F}^{\tau}$. Hence $r_1 \le r_2 : = \operatorname{diam} \overline{U \cap F}^{\tau}$. Let's prove that $r_2 = r_3 := \operatorname{diam} U \cap F$. Clearly, $r_2 \ge r_3$. Fix $\phi ,\psi \in \overline{U \cap F}^{\tau}$. There are nets $(\phi_d)_{d\in D}, (\psi_d)_{d\in D'}$ in $U \cap F$ that converges to $\phi,\psi$ in $\tau$ respectively. Then there exist subnets of $(\phi_d)_{d\in D}, (\psi_d)_{d\in D'}$ that share the same directed set and that their difference is a net converging to $\phi-\psi$ in $\tau$. Because each element of the difference net belongs to the $\tau$-closed ball $r_3B_{E''}$, so does $\phi-\psi$. Hence $r_2 \le r_3 \le r_1$, which implies $\operatorname{diam} U = \operatorname{diam} U \cap F$. Let $\hat x, \hat y \in U \cap F$. Because $J$ is a linear isometry,
$$
1 -\delta < 1- \frac{\delta}{2} < \left \langle \frac{\hat x +\hat y}{2}, f \right \rangle \le \left \| \frac{\hat x +\hat y}{2} \right \| = \left |\frac{x + y}{2} \right | .
$$
By uniform convexity of $E$, we get $|x-y| < \varepsilon$. This completes the proof.
Best Answer
By Kakutani's theorem, $M$ is reflexive if and only if $B_M := \{x\in M | |x| \le 1\}$ is compact in the weak topology $\sigma(M, M')$. Notice that $\sigma(M, M')$ coincides with the subspace topology that $\sigma(E, E')$ induces on $M$. So $M$ is reflexive if and only if $B_M$ is compact in $\sigma(E, E')$.
Also by Kakutani's theorem, we have $B_E$ is compact in $\sigma(E,E')$. A closed subset of a compact set is also compact, so it suffices to show that $B_M$ is closed in $B_E$ w.r.t. $\sigma(E, E')$. This is indeed true because $B_M = M \cap B_E$ where $M$ is convex and closed in norm topology, and thus closed in $\sigma(E, E')$.