The direction $\Longleftarrow$ is immediate from the definition of the weak* topology. The direction $\Longrightarrow$ is what has me confused. The proof goes as follows:
Suppose $\varphi$ is a functional on $X^*$ which is continuous with respect to the weak* topology. Then $\varphi^{-1}(B_1^\mathbb{F})$ is an open neighborhood of $0$ in $X^*$, so there exist $x_1,\ldots,x_n$ and $\epsilon>0$ with $U_{x_1,\ldots,x_n,\epsilon}(0)\subseteq \varphi^{-1}(B_1^\mathbb{C})$ (where $U_{y_1,\ldots,y_m,\delta}(g)\subseteq X^{*}$ denotes a basic open set of radius $\delta>0$ around a functional $g\in X^*$). One can now show that $\ker\hat{x}_1\cap\dots\cap\ker\hat{x}_n\subseteq \ker(\varphi)$, so using the First Isomorphism Theorem, we can factor $\varphi$ through the quotient of $X^*$ by $Z:=\cap_{j=1}^n\ker\hat{x}_j$. Moreover, $X^*\backslash Z$ is of dimension at most $n$, because the mapping $X^*\backslash Z\to\mathbb{F}^n$, $f+Z\mapsto[\hat{x}_1(f)\dots \hat{x}_n(f)]^T$ is injective. So far, so good.
$\textbf{The following is what confuses me}$: The book then claims that it follows that $\varphi=\sum_{j=1}^n\alpha_j\hat{x}_j$ for some scalars $\alpha_1,\ldots,\alpha_n$. How do I see this? I tried showing that the maps $X^*\backslash Z\to\mathbb{F},f+Z\mapsto \hat{x}_i(f)=f(x_i)$ span the dual space of $Y=X^*\backslash Z$, but failed in doing so. Is this even the right approach?
