The question is not directly related to seperabilty, but is more about the "limits" of countability.
There is a known proof by Cantor which establishes that the cardinality of the real numbers is strictly greater than the cardinality of the natural numbers (Which is the same as the cardinality of rational numbers, which are dense in the set of real numbers).
It starts with an assumption that the cardinality of the real open interval $(0,1)$ is $\aleph_0$. Every elements is a sequence $0.a_1a_2a_3...$ the next (greater or equals) is $0.b_1b_2b_3...$ and the list continues and is countable. The digit sequence of the diagonal $0.a_1b_2c_3...$ is still a member of the interval but it is in no way an element of the list, contrary to assumption. Therefore it has greater cardinality and is defined as a continuum.
This is wrong, but very close to the proof I can recall from a course in discrete mathematics.
Now, if a space $X$ is separable there exists a countable subset $J$ (can be ordered as a sequence obviously) which is dense in $X$. if the index of the sequence is a finite "dimensional" tuple $(a_{i_1,i_2,..,i_n})$ it is still countable (For the same reason the rational numbers are countable).
If the index family is in itself a collection of infinite sequences, where each sequence is of natural entries then it is listable in much the same way a digital representation of a number is listable by order.
This may lead to a belief that this is still countable, in analogy for the flawed proof of uncountability of the real numbers.
Even if the set from which individual elements in each index is a finite $N>1$, then it still holds that given that the index is a still an infinite sequence:
$$\aleph_0^{\aleph_0} \geq N^{\aleph_0} = 2^{\aleph_0} > \aleph_0$$
The last inequality is also Cantor's theorem that states that for a set $S$ the cardinality of its power set is strictly greater than that of $S$.
So, given the continuum hypothesis, What can be said about this set? And, given not continuum hypothesis, what can be said? :)