= Solution
<Mazur theorem> says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the <Hahn-Banach separation theorem> gives $f\in X^*$ and a real number $\alpha$ such that
$$
\operatorname{Re}f(x)>\alpha\geq\sup_{y\in C}\operatorname{Re}f(y).
$$
This weakly open separation shows that $x$ is outside the weak closure.
By definition,
$$
x_n\rightharpoonup0
\quad\Longleftrightarrow\quad
f(x_n)\longrightarrow0\quad\text{for every }f\in X^*.
$$
For each $n$, regard $x_n$ as the bounded functional $Jx_n:X^*\to\mathbb F$ given by $Jx_n(f)=f(x_n)$. Pointwise convergence makes the family $(Jx_n)$ pointwise bounded on the Banach space $X^*$. The <Uniform boundedness principle> gives
$$
\sup_n\|x_n\|=\sup_n\|Jx_n\|<\infty.
$$
Suppose first that $x_n\rightharpoonup0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\varepsilon$.
Conversely, if weak convergence fails, some $f\in X^*$ and an infinite subsequence satisfy either $\operatorname{Re}f(x_n)\geq\varepsilon$ throughout or $\operatorname{Re}f(x_n)\leq-\varepsilon$ throughout. Every convex combination from that subsequence then has norm at least $\varepsilon/\|f\|$, contradicting the stated property.
Now let the $x_n\in\ell^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\|x_n\|_p\leq C$. For $N$ distinct terms,
$$
\left\|\frac1N\sum_{j=1}^N x_{n_j}\right\|_p^p
=\frac1{N^p}\sum_{j=1}^N\|x_{n_j}\|_p^p
\leq C^pN^{1-p}\longrightarrow0.
$$
The convex-combination criterion therefore proves $x_n\rightharpoonup0$. This is <weak convergence of bounded disjointly supported sequences in lp>.
The final implication for a commutative unital <C-star algebra>[C*-algebra] is true. By the <Commutative Gelfand--Naimark theorem>, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the <Riesz-Markov-Kakutani representation theorem>; the <dominated convergence theorem> gives
$$
\int_Ka_n\,d\mu\longrightarrow0.
$$
Hence $a_n\rightharpoonup0$.
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