Proof, [Theorem 3.34, Fol16]. Let $x, y \in G$ and suppose that for any irreducible unitary representation $(H, \pi)$ of $G$, $\pi(x) = \pi(y)$. Let $\phi \in S(G)$ be an extreme point of $S(G)$ and $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be its GNS triple, then $(H_{\phi}, \pi_{\phi})$ is irreducible by Proposition 32.3.5. In particular, by Corollary 32.3.4,

By (2) of Theorem 32.4.8, the convex hull of extreme points of $S(G)$ is weak*-dense in $S(G)$. By Theorem 32.4.7, the weak* topology and the topology of uniform convergence on compact subsets of $G$ coincide on $S(G)$. As such, $\psi(x) = \psi(y)$ for all $\psi \in S(G)$. By linearity, $\psi(x) = \psi(y)$ for all $\psi \in P(G)$ as well.

By linearity and Lemma 32.4.5, $(f * g)(x) = (f * g)(y)$ for all $f, g \in C_{c}(G; \complex)$. Using Proposition 31.4.10, let $\angles{\psi_U}_{U \in \cn_G(1_G)}\subset C_{c}(G; \complex)$ be an approximate identity for $L^{1}(G; \complex)$ such that $\supp{\psi_U}\subset U$ for all $U \in \cn_{G}(1_{G})$, then for any $g \in C_{c}(G; \complex)$, $\psi_{U} * g \to g$ uniformly. Therefore $g(x) = g(y)$ for all $g \in C_{c}(G; \complex)$.

Should $x \ne y$, then Urysohn’s Lemma implies that there exists $g \in C_{c}(G; \complex)$ with $g(x) \ne g(y)$. Since this is not the case, $x = y$.$\square$