Why would $\{f_c:x\mapsto c\|c\in [-1,1]\}\subset (C_\mathbb{R}[0,1],||\cdot||)$ be closed in the wk* topology?

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When proving that $(C_\mathbb{R}[0,1],||\cdot||_\infty)$ is not a dual space of any Banach space using Alaoglu's theorem and Krein-Milman's theorem I'm stuck with the conclusion that $S=\{f_c:x\mapsto c|c\in [-1,1]\}$ must be dense in the unit ball in the wk* topology. However, we know nothing about the wk* topology, since we don't know how the (assumed) predual of $C[0,1]$ acts on $C[0,1]$. So while it is clear that $S$ is not dense in the unit ball in the norm topology on $C[0,1]$, I don't see how to conclude that it can't be dense in the wk* topology.

Any hint would be much appreciated.


Edit: the title of this question is inspired by the fact that I assume we need to prove that $F$ is closed in wk*, so that we can conclude it's not dense. But I don't see how to prove its closed.

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Note that $S$ belongs to the one-dimensional subspace $\{f_c | c \in \mathbb{R}\}$. Then, $C[0,1]$ with the weak-$*$ topology is a locally convex, Hausdorff topological vector space. Hence, its one-dimensional subspaces are closed. Moreover, on finite-dimensional spaces, all Hausdorff vector topologies coincides. This should help to prove the closedness of $S$.