Theorem: Closed Bounded Subset of Real Numbers is Compact

Let $\R$ be the real number line considered as an Euclidean space.
Let $S \subseteq \R$ be a closed and bounded subspace of $\R$.
Then $S$ is compact in $\R$.

Proof:
A closed and bounded subspace $S$ of $\R$ is a closed subspace of some closed real interval $\closedint a b$.
From Closed Subspace of Compact Space is Compact, it suffices to prove that $\closedint a b$ is compact.
Let $\UU$ be any open cover of $\closedint a b$.
Let:
:$G = \set {x \in \R: x \ge a, \closedint a x \text{ is covered by a finite subset of } \UU}$
Let points in $G$ be classified as $\text{good}$ (for $\UU$).
Thus if $z$ is $\text{good}$, then $\closedint a z$ has that finite subcover that is to be demonstrated for the whole of $\closedint a b$.
The aim therefore is to show that $b$ is $\text{good}$.
Let $x$ be $\text{good}$, and $a \le y \le x$.
We have that $\closedint a y \subseteq \closedint a x$.
Thus $\closedint a y$ can be covered with any finite subset of $\UU$ that covers $\closedint a x$.
As $x$ be $\text{good}$, at least one such subset is known to exist.
Thus, by definition, $y$ is $\text{good}$.
Now we show that $G \ne \O$.
At the same time we show that $G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$.
As $\UU$ covers $\closedint a x$, it follows that $a$ must belong to some $U \in \UU$.
So, let $U \in \UU$ be some open set such that $a \in U$.
Since $U$ is open, $\hointr a {a + \delta} \subseteq U$ for some $\delta > 0$.
Hence $\closedint a x \subseteq U$ for all $x \in \hointr a {a + \delta}$.
It follows that all these $x$ are $\text{good}$.
So $G \ne \O$, and:
:$(1): \quad G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$
Now the non-empty set $G$ is either bounded above or it is not.
Suppose $G$ is ''not'' bounded above.
Then there is some $c$ which is $\text{good}$ such that $c > b$.
From our initial observation that if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$, it follows that $b$ is $\text{good}$, and hence the result.
Now suppose $G$ ''is'' bounded above.
By the Continuum Property, $G$ admits a supremum in $\R$.
So let $g = \sup G$.
Suppose that $g > b$.
As $g$ is the ''least'' upper bound for $G$, there exists some $c$ which is $\text{good}$ such that $c > b$.
Again, from our initial observation, it follows that $b$ is $\text{good}$, and hence the result.
{{AimForCont}} that $g \le b$.
From $(1)$ above:
:$\hointr a {a + \delta} \subseteq G$ for some $\delta > 0$
and so $g > a$.
Since $g \in \closedint a b$, $g$ must belong to some $U_0 \in \UU$.
Since $U_0$ is open, there exists some open $\epsilon$-ball $\map {B_\epsilon} g$ of $g$ such that $U_0 \supseteq \map {B_\epsilon} g$.
Since $g > a$, we can arrange that $\epsilon < g - a$.
As $g$ is the ''least'' upper bound, there must be a $\text{good}$ $c$ such that $c > g - \epsilon$.
This means $\closedint a c$ is covered by a finite subset of $\UU$, say $\set {U_1, U_2, \ldots, U_r}$.
Then $\closedint a {g + \dfrac \epsilon 2}$ is covered by $\set {U_1, U_2, \ldots, U_r, U_0}$.
So $g + \dfrac \epsilon 2$ is $\text{good}$, contradicting the fact that $g$ is an upper bound for $G$.
This contradiction implies that $g > b$, and the proof is complete.
{{qed}}