Prove: The weak closure of the unit sphere is the unit ball. I want to prove that in an infinite dimensional normed space $X$, the weak closure of the unit sphere $S=\{ x\in X : \| x \| = 1 \}$ is the unit ball $B=\{ x\in X : \| x \| \leq 1 \}$.
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Here is my attempt with what I know:
I know that the weak closure of $S$ is a subset of $B$ because $B$ is norm closed and convex, so it is weakly closed, and $B$ contains $S$.
But I need to show that $B$ is a subset of the weak closure of $S$.
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for small $\epsilon > 0$, and some $x^*_1,...,x^*_n \in X^*$, I let $U=\{ x : \langle x, x^*_i \rangle < \epsilon , i = 1,...,n \} $
then $U$ is a weak neighbourhood of $0$
What I think I need to show now is that $U$ intersects $S$, but I don't know how.

With the same notations in you question: Notice that if $x_i^*(x) = 0$ for all $i$, then $x \in U$, and therefore the intersection of the kernels $\bigcap_{i=1}^n \mathrm{ker}(x_i^*)$ is in $U$. Since the codimension of $\mathrm{ker}(x^*_i)$ is at most $1$, then the intersection has codimension at most $n$ (exercise: prove this). But since $X$ is infinite dimensional, this means the intersection has an infinite dimension, and in particular contains a line. Since any line going through $0$ intersects $S$, then $U$ intersects $S$.
The same argument can be applied to any point in $B$ (any line going through a point in $B$ intersects $S$), and since you've proved the other inclusion, the weak closure of $S$ is $B$.
