= Solution
By definition of the <weak topology>, $x_n\rightharpoonup x$ exactly when $f(x_n)\to f(x)$ for every $f\in X^*$.
Let $(x_n)$ be bounded and choose a norm-dense sequence $(f_j)$ in $X^*$. Successive subsequences make $f_1(x_n),f_2(x_n),\ldots$ converge, and the <diagonal argument> produces a single subsequence $(y_n)$ on which every $f_j$ converges. Uniform boundedness of $(y_n)$ and norm approximation of an arbitrary $f$ by the $f_j$ show that $f(y_n)$ is Cauchy for every $f\in X^*$. Thus $(y_n)$ is <weakly Cauchy sequence>[weakly Cauchy], and
$$
f(y_{2n}-y_{2n-1})\longrightarrow0
$$
for every $f$, so its difference sequence is <weakly null sequence>[weakly null].
For the countable family $(y_{m,n})_n$, use the weak metric from part a. Delete a finite initial segment from the $m$th sequence so that every remaining term has weak distance less than $1/m$ from zero, and relabel that tail. Enumerate all these tails while preserving the order within each one. For every weak neighbourhood of zero, all terms from sufficiently large $m$ lie inside it, and only finitely many terms from each of the finitely many remaining sequences lie outside it. The resulting enumeration $(z_n)$ is weakly null and contains the relabelled $m$th sequence as a subsequence for every $m$. Equivalently, without relabelling, it contains a tail-subsequence of every original sequence, which is the form used below.
Solved by gpt-5.6-sol high.
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