Here is another way to show the consistency of such a set by a direct symmetric extension approach:
Let $\mathbb{P}$ be the forcing that add Cohen reals (by reals I mean elements of $\omega^\omega$) indexed by $\omega\times\omega$ and let $\mathcal{G}$ be the group of all permutations of $\omega\times\omega$ such that for every $n$, $\pi (n,i) = (n,j)$ for some $j$.
Let $\mathcal{F}$ be the filter on $\mathcal{G}$ generated by $\{H_n \mid n \in \omega\}$ where $H_n$ consists of all $\pi$ such that $\pi (k,i) = (k,i)$ for all $k\le n$, all $i\in\omega$.
Let $x_{k,i}$ be the Cohen reals added in the symmetric extension and $A_k = \{x_{k,i} \mid i \in \omega\}$, then the function $k \mapsto A_k$ is in the symmetric extension and so is $A = \bigcup_{k} k^\smallfrown A_k$, where $x \in k^\smallfrown A_k$ if $x(0) = k $ and there exists $y \in A_k$ such that $x(n+1) = y(n)$ for all $n$.
Now $A$ is not separable, because otherwise there would be a choice function for $k\mapsto A_k$ (which doesn't exists in our symmetric model by construction) but each $A_k$ is separable and dense, therefore, given any $r\in\omega^\omega$, if $r(0) = k$ then I can find a sequence in $k^\smallfrown A_k$ converging to $r$.