= Solution
For a real <locally convex space> $(X,\mathcal P)$, the <continuous dual space> $X^*$ consists of all continuous real-linear maps $X\to\mathbb R$. If $g\in Y^*$ on a subspace $Y$, continuity gives seminorms $p_1,\ldots,p_k\in\mathcal P$ and $C>0$ such that
$$
|g(y)|\leq C\max_i p_i(y)
\qquad(y\in Y).
$$
The right side is a continuous sublinear functional on $X$. The dominated <Hahn-Banach theorem> extends $g$ to a linear $f:X\to\mathbb R$ satisfying the same bound, so $f\in X^*$.
If $Y$ is closed and $x_0\notin Y$, the Hausdorff locally convex quotient $X/Y$ has a continuous seminorm $p$ with $p(x_0+Y)>0$. On $Y+\mathbb Rx_0$, define $g(y+tx_0)=t$ and rescale $p$ so that $|g|\leq p$. Hahn--Banach extends it to $f\in X^*$ with $f|_Y=0$ and $f(x_0)=1$.
The <separation of a point and an open convex set> says that if $C\subseteq X$ is nonempty, open, and convex and $x_0\notin C$, there is $f\in X^*$ such that
$$
f(c)<f(x_0)
\qquad(c\in C).
$$
To prove it, choose $a\in C$, put $U=C-a$, and let $p_U$ be its <Minkowski functional>. For $z=x_0-a\notin U$, $p_U(z)\geq1$. Define $h(tz)=tp_U(z)$ on $\mathbb Rz$; then $h\leq p_U$. Hahn--Banach extends $h$ to $f\leq p_U$. Since $p_U(u)<1$ for $u\in U$, one has $f(c-a)<1\leq f(x_0-a)$, proving the claim.
If $K$ is closed and convex and $x_0\notin K$, a <Hahn-Banach separation theorem> gives $f\in X^*$ and a real number separating $x_0$ from $K$. The corresponding inverse image of an open interval is a weak neighbourhood of $x_0$ disjoint from $K$, so $K$ is closed in $\sigma(X,X^*)$.
The unit sphere $S_X$ of a normed space is norm closed. If $X$ is infinite-dimensional, every basic weak neighbourhood of an interior point $x\in B_X$ constrains only finitely many functionals $f_1,\ldots,f_n$. Their common kernel contains a nonzero $y$. Continuity of $t\mapsto\|x+ty\|$, together with its value below one at $t=0$ and divergence as $|t|\to\infty$, supplies $t$ with $\|x+ty\|=1$. Since all $f_i$ have the same values at $x+ty$ and $x$, every weak neighbourhood of $x$ meets $S_X$. Points outside $B_X$ are separated from it by Hahn--Banach, so the <weak closure of the unit sphere> is exactly $B_X$. In particular, $S_X$ is not weakly closed.
Solved by gpt-5.6-sol high.
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