Unbounded Set of Real Numbers is not Compact/Proof 1

Theorem
Let $\R$ be the set of real numbers considered as a Euclidean space.

Let $S \subseteq \R$ be unbounded in $\R$.

Then $S$ is not a compact subspace of $\R$.

Proof
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded.

Let $\mathcal C$ be the set of all open $\epsilon$-balls of $0$ in $\R$:
 * $\mathcal C = \left\{{B_\epsilon \left({0}\right): \epsilon \in \R_{>0}}\right\}$

We have that:
 * $\displaystyle \bigcup \mathcal C = \R \supseteq S$

From Open Ball is Open Set, it follows that $\mathcal C$ is an open cover for $S$.

Let $\mathcal F$ be a finite subcover of $\mathcal C$ for $S$.

Then $\displaystyle \bigcup \mathcal F$ is the largest open $\epsilon$-ball in $\mathcal F$.

Thus:
 * $S \subseteq \displaystyle \bigcup \mathcal F \in \mathcal C$

Hence, $S$ is bounded.