If $x$ is an element of the closure of $S$, but not $S$ itself, is it true that there exists a sequence in $S$ with limit $x$? If $x$ is an element of $\bar{S}$, but $x$ is not an element of $S$, is it true that there is a sequence $\{ s_n \}$ in $S$ with limit $x$?
My logic is that, because there is a theorem which states that $S$ is closed if and only if all sequences in $S$ that converge have limit in $S$, then if my statement weren't true then one could just remove the point $x$ from $\bar{S}$, and there would be no difference.
So I guess by extension my question is, is the only way to construct $\bar{S}$ to consider all sequences in $S$ with limit outside of $S$, then let that limit point be a part of $\bar{S}$? Or is that even a way at all? Pretty clueless over here. Thanks team. 
 A: The issue with picking "infinitely many balls" is that in most topologies(including metric and norm topologies) that you probably know, there are infinitely many balls around a single point i.e. every point has a neighbourhood basis. This is called first countability, and the same argument shows that if $x \in \bar S$ in a first countable space then there is a sequence in $S$ converging to $x$. Therefore, in the context you are working in, this is true, all very good.
Thus, the question is prompted : what if , there aren't infinitely many balls around a point? More precisely, what if some point does not have a neighbourhood basis i.e. the space is not first countable? The statement : "$x \in \bar S$ implies there is a sequence in $S$ converging to $x$" is not true!

To give an example of such a topology,we have the co-countable topology on $\mathbb R$. The open sets here, are of the form $L\subset \mathbb R$, where either $L$ is empty or $L^c$ is countable. I won't go into why it is not first countable : that is an exercise you may want to attempt. Note that the closed sets, under this topology, are $\mathbb R$ and all countable sets.
Here, take the set $X = \mathbb R \setminus \{0\}$. Now, consider the "closure" of this set. This is the smallest closed set which contains $X$. But then, $\mathbb R$ contains $X$, and any other closed set is countable, so the closure is $\mathbb R$.
But $0$ is not the limit of any sequence in $X$. Why? Well, first let us analyse why something like $\frac 1n$,which is in $X$, does not go to $0$ (but goes in the usual metric topology).
For the general definition of convergence, any open set around $0$ must contain $\frac 1m$ for large enough $m$, only then can we say that $\frac 1m \to 0$. But here, we can take the open set given by $\mathbb R \setminus K$ where $K = \{\frac 1m\}$. This open set does not even intersect the sequence, but contains $0$. Hence, $\frac 1m \not \to 0$!(Already, we are in fishy territory).
But then, we can do what we did with $\frac 1m$ for any sequence : take the open set given by $\mathbb R$ with that sequence removed : this is an open set around $0$ which does not even intersect the given sequence!
Conclusion : No sequence in $S$ converges to $0$.
In fact, it turns out that in this topology, only eventually constant sequences converge, of course to the constant!
This provides a contradiction to a statement that is often taken for granted.

Later, you will learn about "closure" and "sequential closure", where the second is $S$ along with its limit points , and the first is the largest closed set containing $S$. These differ, but it is always true that the latter is contained in the former. 
In weak topologies on Banach spaces, this difference is very emphasised (we keep saying  "weak closure", "weak sequential closure", which are different in that context).
Keep in mind : things you take for granted in $\mathbb R^n$, or even on metric spaces, need not be true all the time!
