Prove that the space of divergent sequences in $(l_{\infty},d_{\infty})$ is open and dense. Is it separable? The problem statements are:
Consider the space $A=\{ \{a_n\}_{n \in \mathbb N} \in l_{\infty} :  \{a_n\}_{n \in \mathbb N} \text { is not convergent }\}$
$a)$ Prove that $A$ is open and dense in $(l_{\infty},d_{\infty})$
$b)$ Decide if $A$ is a separable metric subspace of $(l_{\infty},d_{\infty})$.
My attempt at a solution:
For $a)$, I found it easy to prove density: 
Take $\{a_n\}_{n \in \mathbb N}$ and let $\epsilon>0$, if $\{a_n\}_{n \in \mathbb N} \in A$, then it's immediate that $B(\{a_n\}_{n \in \mathbb N}, \epsilon) \cap A \neq \emptyset$. Now, suppose  $\{a_n\}_{n \in \mathbb N} \in A^c$, i.e.,  $\{a_n\}_{n \in \mathbb N}$ is a convergent sequence. Define $\{y_n\}_{n \in \mathbb N}$ as $y_1=a_1 + \dfrac{\epsilon}{2}$, $y_n=a_n+\dfrac{\epsilon}{4}$ if $n$ is even and $y_n=x_n+\dfrac{\epsilon}{3}$ if $n$ is odd and $n>1$.
We have that $|a_n-y_n|\leq |a_1-y_1|=\dfrac{\epsilon}{2}$, which means $d_{\infty}(\{a_n\}_{n \in \mathbb N},\{y_n\}_{n \in \mathbb N})=\sup_{n \in \mathbb N} |a_n-y_n|=\dfrac{\epsilon}{2}<\epsilon$.
The sequence $\{y_n\}_{n \in \mathbb N}$ can be expressed as $\{y_n\}_{n \in \mathbb N}=\{a_n\}_{n \in \mathbb N}+\{b_n\}_{n \in \mathbb N}$, where $\{b_n\}_{n \in \mathbb N}$ is defined as $b_1=\dfrac{\epsilon}{2}$, $b_n=\dfrac{\epsilon}{4}$ if $n$ is even, and $b_n=\dfrac{\epsilon}{3}$ if $n$ is odd and $n>1$. We want to show that $\{y_n\}_{n \in \mathbb N} \in A$, so suppose not, then $\{y_n\}_{n \in \mathbb N}$ is a convergent sequence, so $\{y_n\}_{n \in \mathbb N}-\{a_n\}_{n \in \mathbb N}=\{b_n\}_{n \in \mathbb N}$ is a convergent sequence, which is clearly absurd. This proves that $B(\{a_n\}_{n \in \mathbb N},\epsilon) \cap A \neq \emptyset$  
Now it remains to prove $A$ is open, but this is equivalent to prove that $A^c$ is closed: Let $x=\{x_n\}_{n \in \mathbb N}$ be a limit point of $A^c$, I want to show that $x \in A^c$, i.e., that $x$ is a convergent sequence. By hypothesis, there is $\{y^n\}_{n \in \mathbb N}$ (I'm using superscripts here because each $y^n$ is a sequence itself). It's sufficient to show that $x$ is a Cauchy sequence. Here I am not so sure if what I've done is correct, I got confused with so many sequences and inequalities:
Let $\epsilon>0$
By hypothesis, $y^n \to x$, so there is $N : \space \forall \space n \geq N$, $sup_{k \in \mathbb N}|y_{k}^n-x_k|=d_{\infty}(y^n,x)<\dfrac{\epsilon}{3}$. 
$|x_k-x_j|\leq |x_k-y_{k}^N|+|y_{k}^N-y_{j}^N|+|y_{j}^N-x_j|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+|y_{k}^N-y_{j}^N|$, Now, the sequence $\{y_{n}^N\}_{n \in \mathbb N} \in A^c$, which means it is a convergent sequence $\implies$ it is a Cauchy sequence. So, there is $n_0 \in \mathbb N$: if $k,j\geq n_0 \implies |y_{k}^N-y_{j}^N|<\dfrac{\epsilon}{3}$ 
Then, for all $k,j \geq n_0$, we have that 
$|x_k-x_j|\leq |x_k-y_{k}^N|+|y_{k}^N-y_{j}^N|+|y_{j}^N-x_j|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}=\epsilon$.
We've proved $x$ is a Cauchy sequence $\implies$ it is convergent $\implies x \in A^c$. Since $A^c$ is closed, then $A$ is open.
For part $b)$ I would appreciate some help. I think that $A$ is not separable, I've tried to choose $S \subset A$ such as $S$ the space of sequences of ones and zeros that are not eventually $1$ or eventually $0$, but I got stuck trying to show by absurd that $S$ is not separable. Any ideas or suggestions for this point? 
 A: a) Openness is usually easier to prove than closedness. If a sequence $(a_n)$ is not Cauchy, then   there  exists $\epsilon>0$ such that for every $N$ there are $n,m\ge N$ with $|a_n-a_m|\ge \epsilon$. 
Now if  $\|b-a\|_\infty<\epsilon/3$, then with the above indices $|b_n-b_m|\ge \epsilon/3$ by the reverse triangle inequality. Thus, the set of all non-Cauchy sequences is open. 
(Note: for real- or complex- valued sequences "non-Cauchy" is equivalent to "non-convergent". But we can imagine a 
metric space $\ell_\infty(X)$ of bounded sequences in some general metric space $X$; then the openness still holds for non-Cauchy sequences, but may fail for non-convergent ones.) 
For density, it suffices to observe  that for every $a\in \ell_\infty$ and every $\epsilon>0$ at least one of two sequences $(a_n)$ and $(a_n+(-1)^n\epsilon)$ fails to converge. 
b) was settled in comments: since "being dense in" is a transitive relation, a dense subset of $\ell_\infty$ cannot contain a countable dense subset.
