= Solution
The <weak topology> $\sigma(X,X^*)$ is the coarsest topology making every <bounded linear functional> continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<\epsilon,\ 1\le j\le m\}$ for finite families in $X^*$.
<Mazur theorem> says that for any <convex> subset $C$ of a real or complex <normed vector space>,
$$
\boxed{\overline C^{\,w}=\overline C^{\,\|\cdot\|}.}
$$
The <norm> closure is <convex>. If $x$ is outside it, the <Hahn-Banach separation theorem> gives a <bounded linear functional> $f$ and a real $a$ with $\operatorname{Re}f(x)>a\ge\sup_{y\in C}\operatorname{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the <norm> closure. The other inclusion follows because the weak topology is weaker than the <norm> topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.
The sequential formulation, <Mazur lemma>, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the <norm> closure of its <convex hull>. Select a finite <convex> combination of the $n$th tail at <norm> distance less than $1/n$ from $x$.
A <weakly bounded set> $D$ satisfies $\sup_{x\in D}|f(x)|<\infty$ for every $f\in X^*$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^*$. The <completeness of the dual space> holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the <unit ball>. The <Uniform boundedness principle> therefore gives $\sup_{x\in D}\|Jx\|<\infty$, and $\|Jx\|=\|x\|$ proves <norm> boundedness.
One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^*:\sup_{x\in D}|f(x)|\le n\}$ cover the <Banach space> $X^*$. The <Baire category theorem> makes some $F_n$ contain a ball $f_0+rB_{X^*}$, after shrinking the radius. For $\|g\|\le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so $\sup_D|g(x)|\le2n$. Scaling and the dual <norm> formula yield $\sup_D\|x\|\le2n/r$. A <weakly compact set> is weakly bounded because each functional has compact, hence bounded, image. Consequently it is <norm> bounded.
For a <Banach space>, let $J:X\to X^{**}$ be its canonical <isometry>. We use two explicitly stated weak-star facts: <Banach-Alaoglu theorem> makes a dual <unit ball> weak-star compact, and <Goldstine theorem> makes $J(B_X)$ weak-star dense in $B_{X^{**}}$. The weak topology on $X$ is carried by $J$ to $\sigma(X^{**},X^*)$ on its image, since their coordinates are the same evaluations $f(x)$.
If $X$ is a <reflexive Banach space>, $J(B_X)=B_{X^{**}}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the <Hausdorff space> $X^{**}$. Goldstine density then forces $J(B_X)=B_{X^{**}}$, which implies $JX=X^{**}$. This proves the <weak compactness characterization of reflexivity>:
$$
\boxed{X\text{ is reflexive}\iff B_X\text{ is weakly compact}.}
$$
Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^*}$ separate points. If $K$ is empty there is nothing to prove; otherwise define
$$
\boxed{d(x,y)=\sum_{n=1}^\infty 2^{-n}\frac{|f_n(x-y)|}{1+|f_n(x-y)|}.}
$$
The summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a <Hausdorff space> is a <homeomorphism>. Thus $d$ induces precisely the weak topology on $K$, the <countable separating family metrizes a weakly compact set> result. No norm-density of the separating family in $X^*$ is claimed or required.
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