Why is $L^{\infty}$ not separable?

To be separable means to have a countable dense subset. Suppose that $(M, d)$ is a metric space and that $U\subseteq M$ be an uncountable subset and $r > 0$. Suppose that for all $x \neq y \in U$, $d(x,y) \ge r$. Let $C$ be any countable subset of $M$. Then $C$ can only meet a countable number of the balls $B_{r/2}(x)$ for $x\in U$. Let $G$ be the union of all $B_{r/2}(x)$ for $x\in U$ that do not meet $C$. $G$ is a nonempty open subset of $M$ that does not meet $C$.

There can be no countable dense subset of $M$.

Consider the case of $\ell^\infty$. For each subset $Q$ of the integers, let $x_Q$ be the sequence that is 1 on $G$ and 0 off of it. The $x_Q$ are uncountable and any two elements of this collection are distance 1 apart. We have just shown that $\ell^\infty$ is not separable.

You can generate a similar construct for $L^\infty$. Consider the uncountable subclass of characteristic functions $\{\chi_{B_r(0)}\}_{r>0}\subseteq L^\infty(\mathbb{R}^n)$. Then each pair of distinct elements in it would be 1 unit distance apart. Ergo there cannot be any countable subset of $L^\infty(\mathbb{R}^n)$ that is dense in it.

With the topology $\|f-g\|_{l^\infty}=\sup_i |f_i-g_i|$ it is easy to show if C is any countable set, and $(f_{ij})_j$ a sequence of elements of $C$, you can take an element $h\in l^\infty$ so that $h_i= 0$ if $|f_{ii}|>1$, $h_i=2$ otherwise, hence $\|f_{j}-h\|>1/2$ for all $j$. Then no ball of radius $1/2$ around $h$ can contain an element of $C$.