= Solution
Suppose $x^{(n)}\in c_0$ converges uniformly to $x$. Choose $n$ with $\|x-x^{(n)}\|_\infty<\varepsilon/2$, then a tail on which $|x_j^{(n)}|<\varepsilon/2$. On this tail $|x_j|<\varepsilon$, proving $x\in c_0$. Linearity is coordinatewise, so \b[$c_0$ is a closed subspace.]
For $a\in\ell^1$, define the <bounded linear functional> $L_a(x)=\sum_ja_jx_j$. Absolute convergence gives $\|L_a\|\leq\|a\|_1$. For each $m$, take $x_j=\overline{a_j}/|a_j|$ when $1\leq j\leq m$ and $a_j\ne0$, and zero elsewhere. This finitely supported sequence belongs to $c_0$ and has norm at most one, giving $\|L_a\|\geq\sum_{j=1}^m|a_j|$. Therefore $\|L_a\|=\|a\|_1$.
Conversely, for $L\in c_0^*$ put $a_j=L(e_j)$. The same finite phase vectors give $\sum_{j=1}^m|a_j|\leq\|L\|$, so $a\in\ell^1$. The finite truncations of every $x\in c_0$ converge in the <supremum norm>, hence continuity gives $L(x)=\lim_m\sum_{j=1}^ma_jx_j=L_a(x)$. This proves the surjective <isometric isomorphism of normed spaces> \b[$\boxed{c_0^*\cong\ell^1}$.]
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