= Solution
\b[For a convex subset of a Banach space, weak closure equals norm closure.] The <Mazur theorem> states that for every <convex set> $C\subseteq X$,
$$
\overline C^{\,w}=\overline C^{\,\|\cdot\|}.
$$
Since every <bounded linear functional> is norm-continuous, the <weak topology> is weaker than the <norm topology>, giving $\overline C^{\,\|\cdot\|}\subseteq\overline C^{\,w}$.
For the reverse inclusion, use this form of the <Hahn-Banach separation theorem>: if $D$ is a nonempty norm-closed <convex set> in a real <normed vector space> and $x\notin D$, there is a continuous real-linear functional $F$ such that
$$
F(x)>\sup_{d\in D}F(d).
$$
Apply it to $D=\overline C^{\,\|\cdot\|}$. In a complex <Banach space>, use the underlying real space; every continuous real-linear $F$ is the real part of the complex <bounded linear functional> $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the <weak topology> of $x$ disjoint from $D$, so $x\notin\overline C^{\,w}$. The empty set case is immediate. This proves the <Mazur theorem>.
\b[A weakly null sequence has disjoint convex blocks converging to zero in norm.] If $x_n\xrightarrow{w}0$, then for every starting index $m$,
$$
0\in\overline{\{x_i:i\ge m\}}^{\,w}\subseteq\overline{\operatorname{conv}\{x_i:i\ge m\}}^{\,w}.
$$
The <Mazur theorem> places $0$ in the norm closure of that tail <convex hull>. Thus a finite <convex combination> of vectors from any tail can have norm less than any prescribed $\varepsilon>0$.
Choose the <convex blocks> recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite <convex combination> from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yields
$$
u_n=\sum_{i=p_n}^{q_n}a_i x_i,\qquad a_i\ge0,\qquad\sum_{i=p_n}^{q_n}a_i=1,\qquad\|u_n\|<2^{-n}.
$$
Therefore $u_n\to0$ in norm. The recursive tail selection is what makes these <convex blocks>, rather than merely unrelated <convex combinations>.
Next suppose $X^*$ is separable and $(x_n)$ is bounded. Use the <canonical embedding into the bidual> $J_X:X\to X^{**}$, where $J_Xx(f)=f(x)$. If $\|x_n\|\le M$, then $J_Xx_n\in M B_{X^{**}}$. The <Banach-Alaoglu theorem> makes this ball compact in $\sigma(X^{**},X^*)$, and <weak-star metrizability of the dual ball> makes it metrizable because its predual $X^*$ is separable. A <compact space> that is a <metric space> is sequentially compact, so some subsequence $(y_n)$ satisfies
$$
J_Xy_n\xrightarrow{w^*}\phi\in X^{**}.
$$
The limit need not belong to $J_XX$; it is precisely the use of the <bidual space> that supplies compactness without assuming reflexivity.
For a <convex block> $u_n=\sum_{i=p_n}^{q_n}a_i y_i$, every $f\in X^*$ satisfies
$$
|f(u_n)-\phi(f)|\le\sum_{i=p_n}^{q_n}a_i|f(y_i)-\phi(f)|
\le\sup_{i\ge p_n}|f(y_i)-\phi(f)|\longrightarrow0.
$$
Both $f(y_n)$ and $f(u_n)$ tend to $\phi(f)$, so
$$
\boxed{y_n-u_n\xrightarrow{w}0.}
$$
This is the useful <convex-block cancellation of a weak-star limit>; it does not require the limit to be a vector in $X$.
\b[The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence.] Let $q:X\to Z=X/Y$ be the <quotient map>, with the <quotient norm>. Since $\|z_n\|\le1$, choose $v_n\in X$ such that
$$
q(v_n)=z_n,\qquad\|v_n\|<\frac32.
$$
This uses the infimum defining the <quotient norm> and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common <weak-star topology>[weak-star] limit in $X^{**}$. Its quotient images $q(y_n)=z_{k_n}$ form a <weakly null sequence>.
Apply the <convex block> construction to $(q(y_n))$ in the <quotient Banach space>. Using the same coefficients on $(y_n)$ gives <convex blocks>
$$
u_n=\sum_{i=p_n}^{q_n}a_i y_i,\qquad\|u_n\|<\frac32,\qquad\|q(u_n)\|\to0.
$$
Set $x_n=y_n-u_n$. The <convex-block cancellation of a weak-star limit> proves $x_n\xrightarrow{w}0$, while
$$
\|x_n\|\le\|y_n\|+\|u_n\|<3,\qquad
\|q(x_n)-z_{k_n}\|=\|q(u_n)\|\longrightarrow0.
$$
Thus the <approximate weakly null lifting through a quotient> is achieved with the required bound:
$$
\boxed{x_n\in3B_X,\quad x_n\xrightarrow{w}0,\quad\|q(x_n)-z_{k_n}\|\to0.}
$$
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