Weak closure of the unit sphere is the closed unit ball

The ball $B$ is convex by the triangle inequality and norm closed because $\mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y\mathrm{\parallel }\mathrm{\mid }\le \mathrm{\parallel }x-y\mathrm{\parallel }$. Thus $B$ is weakly closed and contains $S$, giving ${\bar{S}}^w\subseteq B$.

Fix $x\in B$ and a weak neighborhood of $x$ containing the conditions $\mathrm{\mid }{f}_j(y-x)\mathrm{\mid }<\epsilon$, $1\le j\le m$. There is a nonzero $v$ with all $f_j(v)=0$: choose $m+1$ independent vectors in $X$ by finite induction; their images in ${\mathbb{K}}^m$ are dependent, so a nonzero linear combination of the original vectors lies in the common kernel. This also covers $m=0$.

If $\mathrm{\parallel }x\mathrm{\parallel }=1$, the point $x$ itself works. If $\mathrm{\parallel }x\mathrm{\parallel }<1$, put $T=(2+\mathrm{\parallel }x\mathrm{\parallel })\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$. The real function $h(t)=\mathrm{\parallel }x+tv\mathrm{\parallel }$ on $[0,T]$ satisfies $\mathrm{\mid }h(t)-h(s)\mathrm{\mid }\le \mathrm{\mid }t-s\mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$, $h(0)<1$, and $h(T)\ge T\mathrm{\parallel }v\mathrm{\parallel }-\mathrm{\parallel }x\mathrm{\parallel }=2>1$. The intermediate value theorem gives $t\in [0,T]$ with $h(t)=1$. Then $y=x+tv\in S$ has $f_j(y-x)=0$ for every $j$, so lies in the given neighborhood. Every neighborhood of every $x\in B$ meets $S$, proving $B\subseteq {\bar{S}}^w$ and equality. $\mathrm{\square }$