Let $e_n$ be the $n$-th canonical vector in $C_0$. Then $(e_n)_n$ is a Schauder basis for $C_0$. Indeed, for any $a \in C_0$ we have $a = \sum_{n=1}^\infty a_ne_n$ because
$$\left\|a-\sum_{i=1}^na_ie_i\right\|_\infty = \sup_{i\ge n} |a_i| \xrightarrow{n\to\infty} 0$$
For uniqueness, assume $a = \sum_{n=1}^\infty b_ne_n$. For any $j \in \mathbb{N}$ we have
$$|b_j - a_j| \le \sup_{n\in\mathbb{N}}|b_j - a_j| = \left\|\sum_{i=1}^\infty (b_i-a_i)e_i\right\|_\infty = \left\|\sum_{i=1}^\infty b_ie_i-\sum_{i=1}^\infty a_ie_i\right\|_\infty = \|a-a\|_\infty = 0$$
so $a_j = b_j$.