= Solution
The <weak topology> $\sigma(E,E^*)$ is the coarsest topology making every <continuous linear functional> in $E^*$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\varepsilon$, and a <sequence> converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The <Hahn-Banach theorem> makes the dual separate points, so the <weak topology> is Hausdorff.
Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the <Eberlein-Šmulian theorem>. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge1\}$. It is a <separable Banach space>. The <Hahn-Banach theorem> separates points outside a closed <vector subspace> from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.
Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose <norming functionals> $f_j\in Y^*$ with $\|f_j\|=1$ and $f_j(u_j)=1$. They separate points: for $y\ne0$, a sufficiently close $u_j$ to $y/\|y\|$ gives $|f_j(y)|>0$. This is a <countable norming family>. Extend the functionals to $E$ by the <Hahn-Banach theorem>. The evaluation map
$$
C\longrightarrow\mathbb F^{\mathbb N},\qquad y\longmapsto(f_j(y))_{j\ge1}
$$
is weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a <homeomorphism> onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by $\sum_j2^{-j}\min(1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact <metric spaces> are sequentially compact, giving \b[a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$]. If $Y=\{0\}$ the result is immediate. We did not assert that a separable <Banach space> has a separable dual or that its weak topology is generally metrizable.
Back to article page
