Weak and norm topologies agree iff finite dimensional

If $e_1,\dots ,e_d$ is a finite basis, its coordinate functionals $f_j$ are bounded by the boundedness of the inverse coordinate map. Moreover $\mathrm{\parallel }v\mathrm{\parallel }\le {\sum }_j\mathrm{\mid }{f}_j(v)\mathrm{\mid }\mathrm{\parallel }{e}_j\mathrm{\parallel }$. For $r>0$, the finite conditions $\mathrm{\mid }{f}_j(v)\mathrm{\mid }<r\mathrm{/}(1+{\sum }_j\mathrm{\parallel }{e}_j\mathrm{\parallel })$ imply $\mathrm{\parallel }v\mathrm{\parallel }<r$. Thus every norm ball about zero contains a weak neighborhood. Translating gives this at every point; weak-open sets are already norm open because their defining functionals are bounded. The topologies coincide. For $d=0$, $X$ is a singleton and the assertion holds directly.

Suppose instead $X$ is infinite dimensional. Given finitely many $f_1,\dots ,f_m$, choose $m+1$ independent vectors by finite induction. Their images in ${\mathbb{K}}^m$ are dependent; the resulting nontrivial combination gives a nonzero common-kernel vector $v$. Every real multiple $tv$ satisfies every zero-centered finite disk condition, and $\mathrm{\parallel }tv\mathrm{\parallel }=\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ is unbounded. By F1 every weak zero-neighborhood is therefore unbounded. It cannot lie in the norm unit ball, whereas equality of the topologies would make that ball a weak neighborhood. This excludes equality in infinite dimension and completes the equivalence. $\mathrm{\square }$