= Solution
Let $1<p<\infty$ and $q=p/(p-1)$. The dual of $\ell^p$ is $\ell^q$. For every $a\in\ell^q$, $\langle a,e_n\rangle=a_n\to0$, so $e_n\rightharpoonup0$ weakly. A <compact operator> sends a weakly convergent <sequence> to a norm-convergent <sequence> with the corresponding limit. Indeed, any failure of $\|Te_n\|\to0$ would give a <subsequence> bounded away from zero, with a further norm-convergent <subsequence> by <compactness>; weak <continuity> of $T$ forces that <norm> limit to be zero, a contradiction. The condition is therefore \b[necessary].
For failure of sufficiency, partition the positive integers into consecutive blocks
$$
B_k=\{k(k-1)/2+1,\ldots,k(k+1)/2\},\qquad |B_k|=k,
$$
and define the <block averaging compactness counterexample>
$$
\boxed{(Tx)_k=k^{-1/q}\sum_{j\in B_k}x_j.}
$$
By <Hölder's inequality>, $|(Tx)_k|^p\leq\sum_{j\in B_k}|x_j|^p$. Summing over disjoint blocks shows $\|Tx\|_p\leq\|x\|_p$, so $T$ is bounded. If $j\in B_k$, then $Te_j=k^{-1/q}e_k$; as $j\to\infty$, also $k\to\infty$, so $\|Te_j\|_p\to0$.
However, the <vectors> $x^{(k)}=k^{-1/p}\mathbf1_{B_k}$ have <norm> one and satisfy $Tx^{(k)}=e_k$. Distinct output <vectors> have distance $2^{1/p}$, so this bounded <sequence> of inputs has no norm-convergent image <subsequence>. Thus \b[$T$ is not <compact>].
To connect with the hint, define $(Uy)_j=k^{-1/q}y_k$ for $j\in B_k$. Then $U:\ell^q\to\ell^q$ is an isometry, and the dual pairing calculation gives $T=U^*$. The <vectors> $Ue_k$ have disjoint supports and <norm> one. The direct estimates above establish the counterexample without relying only on the hint.
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