I want to prove exactly what is written in the title, using the following idea:
Let $X$ be a separable Banach space. We have that $X$ is isometrically isomorphic to a subspace $M$ of $C(B_{X^\ast})$, where $B_{X^\ast} = \{x^\ast\in X^\ast:\ \|x^\ast\| = 1 \}$ and $C(B_{X^\ast})$ is the space of continuous functions from $B_{X^\ast}$ to $\mathbb{K}$ ( we have $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$).
I want to prove that $M$ is isometrically isomorphic to some subspace of $\ell_\infty$. Doing this it's sufficient to finish the proof, because then $X\cong M\cong $ some subspace of $\ell_\infty$. The first isometric isomorphism is already proven, the second one is the problem. I came up with a solution, but I'm not sure if it's right because the Banach hypothesis was not used (or I just missed where it's used). I appreciate your help, thanks.
My solution: $B_{X^\ast}$ is $w^\ast$-separable so let $(x^\ast_n)_{n\in\mathbb{N}}$ be a $w^\ast$-dense sequence in $B_{X^\ast}$ and define $\Phi:M\to\ell_\infty$ such that $\Phi(f) = (f(x^\ast_1),\ldots,f(x^\ast_n),\ldots)$. It's not hard to see $\Phi$ is well defined, for $|f(x^\ast_n)| \leq \|f\| \cdot \|x^\ast_n\| \leq \|f\|$ for all $n\in\mathbb{N}$, so $\Phi(f)\in \ell_\infty$ for all $f\in C(B_{X^\ast})$.
It's not hard also to prove that $\Phi$ is a linear injection, so the main problem is to prove that $\Phi$ is an isometry. Take any $f\in C(B_{X^\ast})$, we want to prove $\|\Phi(f)\| = \|f\|$, and for this we just need to prove that $\sup_{n\in\mathbb{N}}|f(x^\ast_n)| = \sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)|$ ( for now, all we know is $\sup_{n\in\mathbb{N}}|f(x^\ast_n)| \leq \sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)|$).
First of all, note that $B_{X^\ast}$ is $w^\ast$-compact (by Banach-Alaoglu theorem), then $\sup_{x^\ast\in B_{X^\ast}} |f(x^\ast)| = \max_{x^\ast\in B_{X^\ast}} |f(x^\ast)| = |f(y^\ast)|$ for some $y^\ast \in B_{X^\ast}$. Also, we know $f$ is continuous in $y^\ast$, so for any $\varepsilon > 0$ there is $\delta > 0$ such that $\|x^\ast – y^\ast\| < \delta \implies |f(x^\ast) – f(y^\ast)| < \varepsilon$.
We don't know if there is some $x^\ast_n$ close enough to $y^\ast$, for the separability is about the $w^\ast$ topology, not the norm, so let's work with that. Given any $w^\ast$-neighborhood $V$ of $0$ in $B_{X^\ast}$, we know $V$ is limited, so consider the collection of $w^\ast$-neighborhoods $(\frac{1}{m}V)_{m\in\mathbb{N}}$. Take $m\in\mathbb{N}$ such that $\frac{1}{m} < \delta$, since $(x^\ast_n)_{n\in\mathbb{N}}$ is $w^\ast$-dense in $B_{X^\ast}$, there is some $x^\ast_n$ such that $$x^\ast_n \in y^\ast+\frac{1}{m}V \implies x^\ast_n – y^\ast \in\frac{1}{m}V \subset \delta V \subset \delta B_{X^\ast} \implies \|x^\ast_n – y^\ast\| <\delta .$$
From this we conclude that $|f(x^\ast_n) – f(y^\ast)| < \varepsilon$, but $\varepsilon > 0$ is arbitrary, so there is functionals $x^\ast_n$ such that $f(x^\ast_n)$ is arbitrarily close to $f(y^\ast)$. With this we can conclude that
$$\sup_{n\in\mathbb{N}} |f(x^\ast_n)| = |f(y^\ast)|,$$
and this proves $\Phi$ is a isometry.
PS: There is this thread. But it looks like a different approach.
Best Answer
I have two comments.
I'm not sure that $\Phi$ is surjective. However, since you don't want to prove that $X$ is isometrically isomorphic to $\ell_\infty$ (but to a subspace of $\ell_\infty$), you can replace "linear bijection" by "linear injection".
Every separable set has a countable norming set (Lemma 6.7 here). And any normed space with a countable norming set is isometric to a subspace of $\ell_\infty$ (Corollary 6.8 here). Thus, the completeness (that seems to be not used in your argument) is not really needed.
So, your solution looks fine to me.