Can the closure of a countable set be characterized sequentially? Suppose that I have a countable subset $S \subset X$, where $(X, \tau)$ is a topological space that is NOT first countable (so that convergence is characterized by nets and not sequences). I'm interested in the closure $cl(S)$, which is defined as the collection of all limit points of $S$, where limit points of $S$ are defined as all elements $x \in X$ such that for every open set of $X$ containing $x$, there exists some element of $A$ within that open set (but does not equal $x$ itself). 
My question is then the following: Because the set $S$ is itself countable, can we say that $cl(S)$ is equal to the set of limit points of all sequences of elements in $S$ ? I understand when considering a potential limit point of $A$, there could be an uncountable many open sets that elements in $A$ must fall within, but on the other hand, there are only a countable many elements of $A$ in the first place. On which side of things does this fall?
Thanks!

No. A classical counterexample is the space described in this question. (Arens space). The closure of all points $\mathbb{N} \times \mathbb{N}\setminus \{(0,0\}$ is all of the space, but no sequence from it converges to $(0,0)$.
See also this blog post
