Insert paragraph about the importance of compactness
Let S \subset \mathbb{R}. We say S is connected if there do not exist disjoint open sets U,V \subset \mathbb{R} such that S \cap U and S \cap V are nonempty and
S = (S \cap U) \cup (S \cap V).
The motivation behind this definition is that a set A space is connected if, and only if, it cannot be divided in two nonempty, open and disjoint subsets.
A set of real numbers is connected if and only if it is an interval.
(\Rightarrow) We first want to show every connected set is an interval. Let A \subset \mathbb{R} not be an interval. Then there exists a,b \in A and c \not \in A such that a < c < b by the definition of an interval. Then the open sets U = (-\infty,c) and (c,\infty) separate A. Thus every set that is not an interval is disconnected, and every connected set is an interval.
(\Leftarrow) We now want to show every interval is connected. Let I \in \mathbb{R} be an interval. Towards a contradiction say I is disconnected. Then there exist open sets U,V that separate I. Choose a \in I \cap U and b \in I \cap V. Without loss of generality, take a < b. Let
c = \sup (U \cap [a,b]).
We will show that a < c < b, and that c \not \in I, therefore I is not an interval. We start by showing c is between a and b.
If there exist x \in U such that a \leq x < b, then [x,x+\varepsilon) \subset U for some \varepsilon > 0 since U is open. So x \not = \sup (U \cap [a,b]) Thus c \not = a and if a < c < b, c \not \in U.
If there exist y \in V such that a < y \leq b, then (y-\varepsilon,y] \subset V for some \varepsilon > 0 since V is open. So (y - \varepsilon,y] is disjoint from U, meaning y \not = \sup(U \cup [a,b]). It follows that c \not = b and c \not \in U \cap V. Thus a < c < b and c \not \in I, meaning I is not an interval, and thus a contradiction. Thus I is connected.
So for real numbers, connectedness is the same property as being an interval, which makes it pretty boring.