= Solution
The <weak topology> $\sigma(X,X^*)$ is the coarsest <topology> making every $f\in X^*$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\varepsilon$. The <weak-star topology> $\sigma(X^*,X)$ is the coarsest <topology> making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.
The <canonical embedding into the bidual> is
$$
J_Xx\in X^{**},\qquad(J_Xx)(f)=f(x).
$$
It is linear and $\|J_Xx\|\leq\|x\|$. For $x\ne0$, the <Hahn-Banach theorem> supplies a <norming functional> $f$ with $\|f\|=1$ and $f(x)=\|x\|$, proving equality. Thus $J_X$ is an <isometric embedding>. The coordinates defining the <weak-star topology> on $X^{**}$, restricted to $J_X(X)$, are precisely $J_Xx\mapsto f(x)$ for $f\in X^*$. These are exactly the coordinates defining the <weak topology> on $X$, so $J_X$ is also a <topological embedding> for these <topologies>.
A <normed vector space> is a <reflexive space> if $J_X$ is onto $X^{**}$. Such a space is automatically a <Banach space>, since it is isometric to the complete <continuous dual space> $X^{**}$. The <Banach-Alaoglu theorem> states that the closed unit ball of a <continuous dual space> is compact for its <weak-star topology>. The <Goldstine theorem> states that $J_X(B_X)$ is dense in the <weak-star topology> in $B_{X^{**}}$. Consequently
$$
\boxed{X\text{ is reflexive}\iff B_X\text{ is weakly compact}.}
$$
If $J_X$ is onto, its weak-to-weak-star <homeomorphism> carries $B_X$ onto the <Banach-Alaoglu theorem>'s compact ball $B_{X^{**}}$. Conversely, if $B_X$ is <weakly compact>, $J_X(B_X)$ is compact and hence closed in the <Hausdorff> <weak-star topology>. Its density from the <Goldstine theorem> makes it all of $B_{X^{**}}$, and scaling gives surjectivity of $J_X$.
Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the <Hahn-Banach theorem>, choose <norming functionals> $f_n$ with $\|f_n\|=1$ and $f_n(u_n)=1$. These <continuous linear functionals> separate points: for $x\ne0$, choose $\|u_n-x/\|x\|\|<1/2$ to get $f_n(x)>\|x\|/2$. Define
$$
d(x,x')=\sum_{n\geq1}\min\{2^{-n},|f_n(x-x')|\}.
$$
The <series> converges uniformly, separates distinct points, and obeys the <triangle inequality> because $\min(c,a+b)\leq\min(c,a)+\min(c,b)$. Thus $d$ is a <metric>. Its <topology> is coarser than the <weak topology>: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most $\sum_{n>N}2^{-n}=2^{-N}$. On a <weakly compact set> $K$, the identity $(K,\text{weak})\to(K,d)$ is therefore a continuous bijection from a <compact space> into a <Hausdorff space>, hence a <homeomorphism>. This proves that \b[$K$ is metrizable in its <weak topology>], an instance of <countable separating family metrizes a weakly compact set>.
If $X$ is also a <reflexive space>, its closed unit ball is <weakly compact> and therefore metrizable by this construction. Compactness in a <metric space> implies <sequential compactness>, so every sequence in $B_X$ has a <weakly convergent> subsequence. Scaling proves the claim for every bounded sequence.
For the converse, assume every bounded sequence in $X$ has a <weakly convergent> subsequence. Use the same <metric> $d$ on $B_X$. A <weak limit> of points of $B_X$ remains in $B_X$: $|f(x)|\leq1$ for every $\|f\|\leq1$, and the <Hahn-Banach theorem> recovers $\|x\|\leq1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.
We must still recover the entire <weak topology>, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the <weak topology> $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of <metric> limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies <weak convergence>. Every $f\in X^*$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a <metric space>. The <weak topology> and $d$-topology on $B_X$ coincide. The ball is <weakly compact>, so the criterion above proves
$$
\boxed{X\text{ separable}:\quad X\text{ reflexive}\iff\text{every bounded sequence has a weakly convergent subsequence}.}
$$
The zero space satisfies all statements directly. The converse did not assume completeness, nor use the full <Eberlein-Šmulian theorem>.
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