Proof that if $S$ is closed then $S^c$ is open from Pugh's text Recall $S$ is closed if it contains all the limit points and $S$ is open if the metric $d(p,q) < r$ around every $q \in S$ for some $r>0$
Pugh's real analysis provides a proof as follows:

Suppose $S$ is closed, claim $S^c$ is open. 
Take any $p \in S^c$. 
If there fails to exist an $r > 0$ such that $d(p,q) < r \Rightarrow q
 \in S^c$, then for each $r = 1/n, n = 1,2, \ldots$, there exists a
  point $p_n \in S$ such that $d(p,p_n) < 1/n$. 
This sequence in $S$ converges to $p \in S^c$, contrary to closedness
  of $S$. 
So there exists a $r>0$ such that $d(p,q) < r \Rightarrow q \in
 S^c$ and $S^c$ is an open set.

I feel like this proof is really difficult to interpret.
Can someone please explain to me what is meant by: 


*

*If there fails to exist an $r > 0$ such that $d(p,q) < r \Rightarrow q
 \in S^c$, then for each $r = 1/n, n = 1,2, \ldots$, there exists a
point $p_n \in S$ such that $d(p,p_n) < 1/n$. 

*And this sequence in $S$ converges to $p \in S^c$, contrary to closedness
of $S$. 
In 1. we said nothing converges in $S^c$, then in 2. we say something do converge in $S^c$. I am very confused!
 A: I'm not familiar with Pugh's textbook, but presumably this is all happening in a general metric space $X$. The set $S^c$ (or any set in $X$, for that matter) is open if and only if the following is true:

For every point $p \in S^c$, there exists a real number $r_p > 0$ such that the set $B_{r_p}(p) = \{x \in X : d(p,x) < r_p\}$ is contained in $S^c$.

(By the way, each set $B_{r}(p) = \{x \in X : d(p,x) < r\}$ with $r > 0$ is called the open ball of radius $r_p$ centered at $p$.)
So, if we assume that $S^c$ is not open, then there exists some point $p$ in $S^c$ such that for every $r > 0$ the ball $B_r(p)$ is not contained in $S^c$, which means at least one point $x$ in $B_r(p)$ lies in the complement $X \setminus S^c = (S^c)^c = S$ of $S^c$. Since $1/n > 0$ for each $n \in \mathbb{N}$, we may set $r = 1/n$ in the last sentence to see that for every natural number $n$ there exists $p_n \in B_{1/n}(p)$ such that $p_n \in S$. (Here $p_n$ denotes the $x$ in $B_{r}(p)$ for $r = 1/n$.)
Now then, it sounds like Pugh says that a set is closed if and only if it contains all of its limit points, which in this case means that if $\{s_n\}$ is a sequence of elements of $S$ converging to some element $x$ of $X$ then $x \in S$. We have just shown that $\{p_n\}$ is a sequence in $S$ such that $d(p_n,p) < 1/n$ (by the definition of $B_{1/n}(p)$) for all $n \in \mathbb{N}$. Hence $\{p_n\}$ is a sequence in $S$ converging to $p \not\in S$, which contradicts our assumption that $S$ is closed.
A: If there fails to exist some such  $r$, then there is a sequence $(p_n)_{n\in N}$ of points of $S$ converging to $p,$ implying $p\in S,$ because $S$ contains its limits. But $p\not \in S.$ This  means that the non-existence of some such $r$ is absurd, so some such $r$ must exist.
