Denote by $\mathbb{R}^\mathbb{N}$ the space of all real sequences. Endow this space with the uniform topology induced by the metric
$$ d_u(x,y):=\sup_{k \in \mathbb{N}}\{ \min\{|x_k-y_k|,1\}, $$ where $x,y \in \mathbb{R}^\mathbb{N}$. It is known that this space is not separable.
Define now an equivalence relation on $\mathbb{R}^\mathbb{N}$ as follows. We say that two sequences $x=(x_k)_{k \in \mathbb{N}}$ and $y=(y_k)_{k \in \mathbb{N}}$ are equivalent iff
$$ \limsup_{k \to \infty} |x_k-y_k|=0 $$ and we write $x \sim y$. What can we say about the topological properties of the quotient $\mathbb{R}^\mathbb{N}/\sim$ endowed with the quotient topology? Could the quotient be separable? I suppose not, but I would like to have a proof. Thanks in advance.
Define a pseudometric $d_l$ on $\mathbb R^{\mathbb N}$: $$ d_l(x,y) = \limsup_{k} |x_k-y_k| $$ It is easy to see that $d_l$ respects $\sim$, so it becomes an actual metric on $\mathbb R^{\mathbb N}$.
Furthermore, this metric generates the quotient topology: Any open ball according to $d_l$ is open in the quotient topology, and conversely any open set in the quotient topology contains a $d_l$-ball around each of its points. (The verification of these properties is somewhat simpler once we notice that the spaces are actually normed vector spaces rather than general metric spaces, so many things need to be proved only around $0$).
With these preliminaries, we can diagonalize to show that the quotient is not separable. If $\sigma$ is any function $\mathbb N \to \mathbb R^{\mathbb N}$, then setting $$ x_{2^k(1+m)} = 1+\sigma(k)_{2^k(1+m)} $$ will guarantee that $d_l(x,\sigma(n)) \ge 1$ for every $n$, and therefore the range of $\sigma$ is not dense.