In the proof of Lemma $15$, the author claimed that if there exists constants $C>0$ and $D>0$ such that
$$\begin{array}{lcl}C \sup \{ | \sum_{i=1}^k{\alpha(i) f(x_i) | \colon \| f \|_{\infty}^1 \leqslant 1, f(0)=0} \} &\leqslant &\sup \{ |\sum_{i=1}^k{\alpha(i) f(x_i)| : \| f \|_{Lip} \leq 1, f(0)=0} \} \\ &\leqslant & D \sup \{ |\sum_{i=1}^k{\alpha(i) f(x_i)}| \colon \| f \|_{\infty}^1 \leqslant 1, f(0)=0\}\end{array}$$
then the map $T(\delta_x)(f) = f(x)$ is an isomorphism.
Why the inequalities above implies isomorphism?
It implies that $C\|x\|\leqslant \|Tx\|$ holds true which means that $T$ is injective and has closed range. Note that it is enough to check this only on a dense subspace and ${\rm span}\{\delta_x\colon x\in X\}$ is such a subspace.