= Solution
The compact set $S(B_{c_0})$ is bounded, so $S$ is a <bounded linear operator>. For each $g\in Y^*$, the <continuous linear functional> $g$ attains its maximum on that <compact set>. Symmetry of $S(B_{c_0})$ implies that maximum is $\|S^*g\|$, so $S^*g$ attains its <norm> on $B_{c_0}$. By <norm attainment on the vanishing sequence space>,
$$
S^*(Y^*)\subseteq c_{00}\subset\ell_1.
$$
Regard the <Banach-space adjoint> $S^*$ as a <bounded linear operator> from the <Banach space> $Y^*$ into $c_{00}$ with its inherited $\ell_1$ <norm>. The <completeness of the dual space> holds even if $Y$ is incomplete. The preceding <Baire category theorem> argument shows that $S^*$ has finite rank.
If $S^*(Y^*)=\operatorname{span}\{f_1,\ldots,f_r\}$, set $N=\bigcap_{j=1}^r\ker f_j$. For $x\in N$ and every $g\in Y^*$, $g(Sx)=(S^*g)(x)=0$. The <Hahn-Banach theorem> implies that <continuous linear functionals> separate points of $Y$, so $Sx=0$. Thus $N\subseteq\ker S$, and
$$
\boxed{\operatorname{rank}S\leq\operatorname{codim}N\leq r<\infty.}
$$
This proves <compact closed-ball image forces finite rank on c0>.
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