A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded

If $K$ is compact then $K$ is closed and bounded, which is [L2]; this is the forward implication.

For the backward implication assume $K$ is closed and bounded. If $K = \varnothing$ then every open cover of $K$ admits the empty subfamily as a finite subcover, so $K$ is compact.

Assume moreover $K \ne \varnothing$; fix $s \in K$ and, by [L5], reals $\ell, u$ with $\ell \le y \le u$ for every $y \in K$. Then $\ell \le s \le u$, so $\ell \le u$, and $K \subseteq [\ell,u]$ by [L6].

Let $\mathcal{U}$ be an open cover of $K$ and put $\mathcal{W} := \mathcal{U} \cup \{\mathbb{R} \setminus K\}$. Every member of $\mathcal{W}$ is open, since $\mathbb{R} \setminus K$ is open by [L4], and $\mathcal{W}$ covers $[\ell,u]$: a point of $[\ell,u]$ either lies in $K$, hence in some member of $\mathcal{U}$, or lies outside $K$, hence in $\mathbb{R} \setminus K$.

By [L3] the interval $[\ell,u]$ is compact, so some finite subfamily $\{W_0, \dots, W_p\}$ of $\mathcal{W}$ covers $[\ell,u]$, where the case of an empty subfamily is possible only when $[\ell,u] = \varnothing$, which is excluded by $\ell \le u$. Put $\mathcal{V} := \{\, W_i : W_i \in \mathcal{U} \,\}$, a finite subfamily of $\mathcal{U}$. Then $K \subseteq \bigcup \mathcal{V}$: a point $y \in K \subseteq [\ell,u]$ lies in some $W_i$, and $W_i$ cannot be a member of $\mathcal{W}$ outside $\mathcal{U}$, because the only such member is $\mathbb{R} \setminus K$ and $y \in K$; so $W_i \in \mathcal{U}$ and $W_i \in \mathcal{V}$.

Every open cover of a nonempty closed bounded $K$ therefore has a finite subcover, so such a $K$ is compact; together with the empty case of step 1.2 this proves the backward implication, and step 1.1 is the forward one.