Proof. Let $\seq{x_n}\subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let

Since $\real^{N}$ is separable, $T_{N}(S)$ is separable by Proposition 7.1.2. Thus there exists $\bracs{\phi_{N, k}}_{k = 1}^{\infty} \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^{\infty}$ is dense in $T_{N}(S)$.

Let $\phi \in S$, then for each $N \in \natp$, there exists $k_{N} \in \natp$ such that for each $1 \le n \le N$,

Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E}\to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N}\to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp}\subset S$ is uniformly equicontinuous, $\phi_{N, k_N}\to \phi$ in the weak*-topology by Proposition 8.11.7.$\square$