A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact

For claim 1, let $(X, \mathcal{T})$ be compact, let $F \subseteq X$ be closed and let $\mathcal{U}$ be a family of open subsets of $X$ with $F \subseteq \bigcup \mathcal{U}$; put $\mathcal{W} := \mathcal{U} \cup \{\, X \setminus F \,\}$, a family of open subsets of $X$ with $\bigcup \mathcal{W} = X$, since every point outside $F$ lies in $X \setminus F$ and every point of $F$ lies in some member of $\mathcal{U}$.

For claim 2, let $n \in \mathbb{N}$, let $K_0, \dots, K_n$ be compact subsets of $X$, put $K := K_0 \cup \dots \cup K_n$ and let $\mathcal{U}$ be a family of open subsets of $X$ with $K \subseteq \bigcup \mathcal{U}$; then $K_m \subseteq \bigcup \mathcal{U}$ for every $m \le n$, so by [L2] the set $T_m$ of finite subfamilies of $\mathcal{U}$ whose union contains $K_m$ is nonempty, the empty subfamily belonging to it when $K_m = \varnothing$.

If $X = \varnothing$ then $F = \varnothing$ and the second alternative of [L2] holds for $F$; otherwise compactness of $X$ applied to $\mathcal{W}$ gives $n \in \mathbb{N}$ and $W_0, \dots, W_n \in \mathcal{W}$ with $X = W_0 \cup \dots \cup W_n$.

The assignment $m \mapsto T_m$ is a function with domain the natural number $\sigma(n)$ all of whose values are nonempty, so a choice function for its values supplies finite subfamilies $\mathcal{V}_0, \dots, \mathcal{V}_n$ of $\mathcal{U}$ with $K_m \subseteq \bigcup \mathcal{V}_m$ for every $m \le n$.

Assume $F \ne \varnothing$, the case $F = \varnothing$ being settled at step 2.1, and fix $x \in F$; then $x \in W_j$ for some $j \le n$, and $x \notin X \setminus F$, so that $W_j \ne X \setminus F$ and hence $W_j \in \mathcal{U}$. Let $j_0$ be the least $j \le n$ with $W_j \in \mathcal{U}$, which exists by the previous sentence, and put $V_j := W_j$ when $W_j \in \mathcal{U}$ and $V_j := W_{j_0}$ otherwise; then $V_0, \dots, V_n \in \mathcal{U}$, and nothing has been selected, $j_0$ being the least admissible index.

The family $\mathcal{V} := \mathcal{V}_0 \cup \dots \cup \mathcal{V}_n$ is a subfamily of $\mathcal{U}$; it is finite, a union of finitely many listable families being listed by concatenating their lists; and $K = K_0 \cup \dots \cup K_n \subseteq \bigcup \mathcal{V}$, since each $K_m$ lies inside $\bigcup \mathcal{V}_m \subseteq \bigcup \mathcal{V}$. So $\mathcal{V}$ is empty, in which case $K = \varnothing$, or listable as $\{U_0, \dots, U_p\}$ with $K \subseteq U_0 \cup \dots \cup U_p$; by [L2] the set $K$ is a compact subset of $X$, which is claim 2.

$F \subseteq V_0 \cup \dots \cup V_n$: given $y \in F$ there is $j \le n$ with $y \in W_j$, and $y \in F$ forces $W_j \ne X \setminus F$, hence $W_j \in \mathcal{U}$ and $V_j = W_j \ni y$. Since $V_0, \dots, V_n$ are members of $\mathcal{U}$, [L2] gives that $F$ is a compact subset of $X$, the case $F = \varnothing$ having been settled at step 2.1.

Claim 1 is step 4.1 and claim 2 is step 3.2, and the final sentence of claim 2 is the compactness of the empty space, which holds because the empty subfamily of any family covers it.