A nice way to see that $(\ell^\infty,d_\infty)$ is not separable Let $(X,d)$ a metric space such that $\exists A\subseteq X$  uncountable and $\exists \epsilon\gt0$ such that $\forall x,y \in A,x\neq y\Rightarrow d(x,y)\gt\epsilon$. Prove that $X$ is not a separable space and and use this result to show that $(\ell^\infty,d_\infty)$ is not separable.
There is a result that says that in a separable metric space, every uncountable subset contains a point of accumulation. There is a result that says that in a separable metric space, every uncountable subset contains a point of accumulation. I started considering the result, which did not lead to anything. Then, by contradiction, I wanted to prove that $X$ is separable. In order to use that, I could see if $X$ has a countable base or a cover of open sets of $X$ has a countable subcover.
 A: Let me show you another way of proving that $(\ell^{\infty}, d_{\infty})$ is not separable. The proof is specific to $\ell^{\infty}$ but it is a cute trick, nonetheless. 
Assume, to the contrary, that there is a countable dense set $S$ in $\ell^{\infty}$. Enumerate $S$ into a list, i.e. $S=\{x_1, x_2, x_3, …\}$. Since $x_i\in\ell^{\infty}$, we can write $x_{i}=(x_{i1}, x_{i2}, x_{i3}, …)$ for each $i\in\mathbb{N}$. Define a sequence $y=(y_n)$ given by:
$$
y_{n}=\begin{cases}
x_{nn}+1 &  \ \ \text{if } |x_{nn}|\leq 1 \\
0 & \ \ \text{if } |x_{nn}|>1
\end{cases}
$$
Since $|y_{n}|\leq 2$ for each $n\in\mathbb{N}$, we see that $y\in\ell^{\infty}$. By construction, $|y_{n}-x_{nn}|\geq 1$, and so $\lVert y-x_n \rVert_{\infty}\geq 1$ for each $x_n\in S$. So $S$ cannot be dense, because, for example $S$ cannot intersect the open ball of radius $1/2$ around the point $y$. Contradiction.
A: If $X$ is separable, then all uncountable subsets of $X$ contain a point of accumulation.
As a result, $A$ contains a point of accumulation, say $x_0$. Consider $B_d[x_0, \epsilon]$. Then, $\left( B_d[x_0, \epsilon] \setminus \{x_0\}\right) \cap A \neq \varnothing$ (since $x_0$ is a point of accumulation). Hence, there's $y_0 \in A$, $y_0 \neq x_0$, such that $d(x_0, y_0) \le \epsilon$. But that's impossible.
A: Hint: if $D$ is dense in $X$, given $x \in A$, we must have $D \cap B(x, \epsilon/2) \neq \varnothing$. Take a point there and call it $f(x)$. We have a map $f: A \to D$, then. Prove that $f$ is injective, and conclude that $D$ is not countable. Since $D$ was an arbitrary dense set, $X$ is not separable.
