 Solution

The compact set $S(B_{c_0})$ is bounded, so $S$ is a bounded linear operator. For each $g\in Y^{\ast }$, the continuous linear functional $g$ attains its maximum on that compact set. Symmetry of $S(B_{c_0})$ implies that maximum is $\Vert S^{\ast }g\Vert$, so $S^{\ast }g$ attains its norm on $B_{c_0}$. By norm attainment on the vanishing sequence space,Regard the Banach-space adjoint $S^{\ast }$ as a bounded linear operator from the Banach space $Y^{\ast }$ 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 $S^{\ast }$ has finite rank.

If $S^{\ast }(Y^{\ast })=\mathrm{span}\{f_1,\dots ,f_r\}$, set $N={\bigcap }_{j=1}^r\ker f_j$. For $x\in N$ and every $g\in Y^{\ast }$, $g(Sx)=(S^{\ast }g)(x)=0$. The Hahn-Banach theorem implies that continuous linear functionals separate points of $Y$, so $Sx=0$. Thus $N\subseteq \ker S$, andThis proves compact closed-ball image forces finite rank on c0.