Let $ c=\{(a_n) : a_n \in \mathbb{R}, \lim_n a_n = L \operatorname{exists} \} $ be equipped with the sup-norm.
Does $c$ have a Schauder basis?
$ c$ is a separable space . That is why my first thought was that it would have no schauder basis. But, it can of course have. My second thought is that since $ u=\operatorname{span}(1,1,...)$ is algebraically (and topologically) complemented in $c$. (i.e $\quad c= c_0 \oplus u $ ) it can be a candidate as a schauder basis. But I do not know how to approach this problem. Can you give me a direction?
As you remarked it is sufficient to show that $c_0$ has a Schauder basis, because $u$ is finite dimensional and therefore a Schauder basis of $c_0$ induces a Schauder basis for $c$. For $i \in \mathbb{N}$, let $e_i$ be i-th vector of the canonical basis, i.e. $e_i(n) := \delta_{in}$ for all $n \in \mathbb{N}$ where $\delta$ is the Kronecker delta. Let $a \in c_0$ with $a = (a_n)_{n \in \mathbb{N}}$, $\lim_{n\to\infty} a_n = 0$. Then we have $$ a = \sum_{i=0}^{\infty}a_i e_i$$ in the supremum norm, which shows that $\{e_i: i \in \mathbb{N}\}$ is a Schauder basis for $c_0$, and therefore $\{e_i: i \in \mathbb{N}\} \cup \{(1,1,\dots)\}$ is a Schauder basis for $c$: Given $a \in c, a = (a_n)_{n \in \mathbb{N}}$, $\lim_{n\to\infty} a_n = L$ we have $$ a = (L,L,\dots) + a - (L,L,\dots) =L\cdot (1,1,\dots) + \sum_{n=0}^{\infty}(a_i -L)e_i.$$