For any $x \in X$, define the set $\mathcal{F}(X) = \overline{\operatorname{span} \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ $\operatorname{Lip}_0(X)$.
The set $\operatorname{Lip}_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on $\operatorname{Lip}_0(X)$ and dual space of $\mathcal{F}(X)$ is $Lip_0(X)$.
In the proof of theorem $7$, which states that $\mathcal{F}([0,1])$ is not linearly isomorphic to $L_1(\mu)$ for every measure $\mu$, the author states the following:
If there is a measure $\mu$ with $\mathcal{F}([0,1])$ linearly isomorphic to a subspace of $L_1(\mu)$, then there is a continuous linear mapping from $L_{\infty}(\mu)$ onto $Lip_0([0,1])$. Since $L_{\infty}(\mu)$ is a commutative C$^*$-algebra, there exists a compact Hausdorff space $K$ such that $L_{\infty}(\mu)$ is isometric to $C(K)$.
Question: How to prove the above two statements?
For first statement, I think we use adjoint mapping, but I don't know how to apply it here. For the second, it resembles the Stone-Weierstrass theorem.