We want to show ${l}^{\infty}$ not separable.
(whereas usual ${l}^{\infty}$ denotes the set of all bounded sequences $f:\mathbb{N}\rightarrow \mathbb{R}$ with $\sup\limits_{n\in \mathbb{N}}|f(n)|<\infty $ and $||f-g|| = \sup\limits_{n\in \mathbb{N}}|f(n)-g(n)|)$
I have thought of an alternative argument to the problem and couldn't find something similar posted. Does the below argument work?
Claim: Every countable set $C \subseteq {l}^{\infty}$ is not dense in ${l}^{\infty}$ .
Idea: Suppose $C = \{f_1,f_2,...\}$ is a countable subset of ${l}^{\infty}$. Then define $g\in {l}^{\infty}$ using the diagonilzation argument where
\begin{equation} g(n)= \begin{cases} -1 & \text{if } f_n(n)\geq 0\\ 1 & \text{if } f_n(n)< 0 \end{cases} \end{equation}
Clearly $g\in {l}^{\infty}$. Also we see that for each $n$ we have $||f_n-g||\geq 1$, since $|f_n(n)-g(n)|\geq 1$. And we can conclude $C$ is not dense.
