I've recently read the proof of Milman–Pettis theorem in Brezis's book of Functional Analysis.
Let $E$ be a uniformly convex Banach space. Then $E$ is reflexive.
I try to get some feeling of how the author arrived at the proof. Please feel free to give comments on my exposition or add your thoughts.
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 $B_E$ and $B_{E''}$ be the closed unit balls of $E, E''$ respectively. Let $J:E \to E'', x \mapsto \hat x$ the canonical injection. We want to prove that $B_{E''} = J[B_E]$. Because $E$ is complete and $J$ an isometry, $J[B_E]$ is closed in norm topology. It boils down to showing that $B_{E''}$ is the norm closure of $J[B_E]$.
Fix $\varphi \in B_{E''}$ and $\varepsilon>0$. WLOG, we assume $\|\varphi\| = 1$. Let $W := \varphi + \varepsilon B_{E''}$. We want to prove that $J[B_E] \cap W \neq \emptyset$. Because the problem involves $B_E$ and $B_{E''}$, it suggests that we can apply Goldstine theorem. For $f \in E'$ and $\lambda >0$, $$ U:= \{\phi \in B_{E''} \mid | \langle \phi-\varphi, f \rangle| < \lambda \} $$ is an open neighborhood (nbh) of $\varphi$ in $\sigma(E'', E'$). By Goldstine theorem, $J[B_E] \cap U \neq \emptyset$, i.e., there is $x\in B_E$ such that $\hat x \in U$. We will pick $f, \lambda$ such that $\hat x \in W$. Suppose the contradiction that $\hat x \in W^c$. It follows that $\hat x \in U \cap W^c$. To utilize the uniform convexity of $E$, let's make some $y\in B_{E}$ appear. Because the definition of uniform convexity is symmetry, we chose $y$ that has the same properties as $x$, i.e, $\hat y \in U \cap W^c$.
By Banach–Alaoglu theorem, $B_{E''}$ is compact and thus closed in the Hausdorff topology $\sigma(E'', E')$. This implies $W$ is closed and thus $W^c$ is open in $\sigma(E'', E')$. Notice that $U \cap W^c$ is a nbh of $\hat x \in B_{E''}$ in $\sigma(E'', E')$. By Goldstine theorem again, there is $y \in B_E$ such that $\hat y \in U \cap W^c$. To summarize, we get $$ \begin{cases} | \langle \hat x - \varphi, f \rangle| < \lambda &\text{ and } \| \hat x - \varphi\| > \varepsilon \\ | \langle \hat y - \varphi, f \rangle| < \lambda &\text{ and } \| \hat y - \varphi\| > \varepsilon \end{cases}. $$
As mentioned above, it's done if we are able to pick $f, \lambda$ such that we arrive at some contradiction. Let $\delta$ be the modulus of uniform convexity, i.e., $$ \forall (x,y\in B_E) \left [ |x-y| > \varepsilon \implies \left | \frac{x+y}{2} \right | < 1-\delta \right ]. $$
First, $| \langle (\hat x + \hat y) - 2\varphi, f \rangle| \le | \langle \hat x - \varphi, f \rangle| + | \langle \hat y - \varphi, f \rangle| <2\lambda$, so $$ \langle \varphi, f \rangle - \left \langle f,\frac{x + y}{2} \right \rangle \le \left | \left \langle f,\frac{x + y}{2} \right \rangle - \langle \varphi, f \rangle \right| = \left | \left \langle \frac{\hat x + \hat y}{2} - \varphi, f \right \rangle \right| < \lambda. $$
It follows that $$ \langle \varphi, f \rangle - \lambda < \left \langle f, \frac{x + y}{2} \right \rangle \le \|f\| \cdot \left | \frac{x + y}{2} \right |. $$
This implies $$ \left | \frac{x + y}{2} \right | > \frac{\langle \varphi, f \rangle - \lambda}{\|f\|}. $$
To use uniform convexity, we want $f, \lambda$ such that $$ \frac{\langle \varphi, f \rangle - \lambda}{\|f\|} \ge 1- \delta . $$
As such, we pick $f\in E'$ such that $\|f\| = 1$ and $\langle \varphi, f \rangle \ge 1+\lambda -\delta$. Because $\|\varphi\| = 1$, we enforce $\lambda$ such that $\lambda - \delta <0$. To obtain a contradiction, we want $|x-y| > \varepsilon$. However, the conditions $\| \hat x - \varphi\| > \varepsilon$ and $\| \hat y - \varphi\| > \varepsilon$ imply $|x-y| = \|\hat x -\hat y\| \le \| \hat x - \varphi\| + \| \hat y - \varphi\| >2\varepsilon$. This is not what we want. So we will modify the construction of $y$ to fix this.
Let $V := \hat x + \varepsilon B_{E''}$. Similar to $W^c$, $V^c$ is open in $\sigma(E'', E')$. We apply Goldstine theorem on $U \cap V^c$ instead of $U \cap W^c$. However, we need to check that $V^c \cap U \neq\emptyset$. Let $(\varphi_n)$ be a sequence in $E''$ that converges in norm topology to $\varphi$. For $n$ large enough, $\varphi_n \in V^c$ (because $\| \varphi - \hat x \| > \varepsilon$) and $\varphi_n \in U$ (because $|\langle \phi-\varphi, f \rangle| \le \|\phi-\varphi\| \cdot \|f\|$ and $\|f\|=1$). In the end, we obtain a contradiction, $$ \left | \frac{x + y}{2} \right | \ge 1- \delta \quad \text{and} \quad |x-y| = \|\hat x-\hat y\| \ge \varepsilon. $$
This completes the proof.