Recently, I have come across an elegant proof of Milman–Pettis theorem. Surprisingly, I'm able to make this proof even simpler. I'm very happy to share it with you and receive your suggestion.
Let $E$ be a uniformly convex Banach space. Then $E$ is reflexive.
PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, then I will happily accept theirs.
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.