TITLE: structure theorem for Banach spaces
QUESTION [6 upvotes]: The following is a theorem in the Banach Algebra Techniques in Operator Theory by Douglas:

Here are my questions: 

Could one come up with some reference (or proof) regarding the remark right after the proof:


if $\mathscr{X}$ is separable, then $X$ can be taken into the closed unit interval $[0,1]$?


Why does the remark say that the canonical $X$ associated with $\mathscr{X}$ is absent? Isn't it $(\mathscr{X}^*)_1$ (the closed unit ball in $\mathscr{X}^*$)?

REPLY [2 votes]: This is the Banach–Mazur theorem. 
When $X$ is separable, the unit ball $B_{X^*}$ of $X^*$ is metrisable in the weak*-topology. Every compact metric space is a continuous image of the Cantor set $\Delta$. Thus $X$ can be regarded as a subspace of $C(B_{X^{*}})$ which in turn is a subspace of $C(\Delta)$ via $$\iota(f) = f\circ \varphi\quad \big(f\in C(B_{X^*})\big)$$ where $\varphi \colon \Delta \to B_{X^*}$ is a continuous surjection. Now it is enough to embed $C(\Delta)$ isometrically into $C[0,1]$. Usually people use the Borsuk(–Dugundji) theorem to find such an embedding. 
This theorem asserts that if $K$ is a compact Hausdorff space and $L$ is a closed, metrisable subspace of $K$ then there exists a norm-one operator $T\colon C(L)\to C(K)$ such that $(Tf)|_L = f$ for all $f\in C(L)$.
There are many ways to embed $C(\Delta)$ into $C[0,1]$ and there is no canonical surjection $\varphi \colon \Delta\to B_{X^*}$ either so that is why the adjective canonical is omitted.