I'm doing Ex 3.20.2 in Brezis's book of Functional Analysis.
There is an isometry from a separable normed space $E$ into $\ell^\infty$.
My solution is different from the author's. Could you have a check on my attempt?
I posted my proof as an answer below, so I can accept my own answer and thus remove my question from unanswered list.
Let $G$ be a countable dense subset of $E$. For each $x \in E$, we define $f_x:\mathbb Rx \to \mathbb R, tx \mapsto t |x|$. Then $f_x \in (\mathbb Rx)'$, $\langle f_x, x\rangle = |x|$, and $\|f_x\|=1$. By Hahn-Banach theorem, $f_x$ admits an extension $g_x \in E'$ such that $\|g_x\| = 1$.
For each $x \in E$, there is a sequence $(x_n) \subseteq G$ such that $x_n \to x$. Then we define $T:E \to \ell^\infty$ by mapping $x \in E$ to the sequence $(\langle g_{x_n}, x \rangle)_n$. We have $$ \sup_{n} |\langle g_{x_n}, x \rangle| \le \sup_n \|g_{x_n} \| \cdot |x| = |x| < \infty. $$
Then $T(x)$ indeed belongs to $\ell^\infty$. Also, $$ \begin{aligned} |\langle g_{x_n}, x \rangle| &= |\langle g_{x_n}, x-x_n \rangle + \langle g_{x_n}, x_n \rangle| && \ge |\langle g_{x_n}, x_n \rangle| - |\langle g_{x_n}, x-x_n \rangle| \\ &= |x_n| - |\langle g_{x_n}, x-x_n \rangle| &&\ge |x_n| - |x-x_n|. \end{aligned} $$
It follows that $$ \lim_n |\langle g_{x_n}, x \rangle| \ge \lim_n |x_n| - \lim_n |x-x_n| = |x|. $$
This implies $$ \sup_{n} |\langle g_{x_n}, x \rangle| = |x|. $$
Then $T$ is the required isometry.