A closed and bounded subspace $S$ of $\R$ is a closed subspace of some closed real interval $\left[{a \,.\,.\, b}\right]$. From Closed Subspace of Compact Space is Compact, it suffices to prove that $\left[{a \,.\,.\, b}\right]$ is compact. Let $\mathcal U$ be any open cover of $\left[{a \,.\,.\,b}\right]$. Let: : $G = \left\{{x \in \R: x \ge a, \left[{a \,.\,.\, x}\right] \text{ is covered by a finite subset of } \mathcal U}\right\}$ Let points in $G$ be classified as $\text{good}$ (for $\mathcal U$). Thus if $z$ is $\text{good}$, then $\left[{a \,.\,.\, z}\right]$ has that finite subcover that is to be demonstrated for the whole of $\left[{a \,.\,.\, b}\right]$. 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 $\left[{a \,.\,.\, y}\right] \subseteq \left[{a \,.\,.\, x}\right]$. Thus $\left[{a \,.\,.\, y}\right]$ can be covered with any finite subset of $\mathcal U$ that covers $\left[{a \,.\,.\, x}\right]$. 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 $G \ne \varnothing$. At the same time we show that $G \supseteq \left[{a \,.\,.\, a + \delta}\right]$ for some $\delta > 0$. As $\mathcal U$ covers $\left[{a \,.\,.\, x}\right]$, it follows that $a$ must belong to some $U \in \mathcal U$. So, let $U \in \mathcal U$ be some open set such that $a \in U$. Since $U$ is open, $\left[{a \,.\,.\, a + \delta}\right) \subseteq U$ for some $\delta > 0$. Hence $\left[{a \,.\,.\, x}\right] \subseteq U$ for all $x \in \left[{a \,.\,.\, a + \delta}\right)$. It follows that all these $x$ are $\text{good}$. So $G \ne \varnothing$, and: : $(1): \quad G \supseteq \left[{a \,.\,.\, a + \delta}\right]$ 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 $g = \sup G$. Let $g > b$. 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. Aiming for a contradiction, assume that $g \le b$. From $(1)$ above: : $\left[{a \,.\,.\, a + \delta}\right) \subseteq G$ for some $\delta > 0$ and so $g > a$. Since $g \in \left[{a \,.\,.\, b}\right]$, $g$ must belong to some $U_0 \in \mathcal U$. Since $U_0$ is open, there exists some open $\epsilon$-ball $B_\epsilon \left({g}\right)$ of $g$ such that $U_0 \supseteq B_\epsilon \left({g}\right)$. 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 $\left[{a \,.\,.\, c}\right]$ is covered by a finite subset of $\mathcal U$, say $\left\{{U_1, U_2, \ldots, U_r}\right\}$. Then $\left[{a \,.\,.\, g + \dfrac \epsilon 2}\right]$ is covered by $\left\{{U_1, U_2, \ldots, U_r, U_0}\right\}$. 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}}