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 $K:=K_0\cup \cdots \cup K_n$ 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 $K_m=\mathrm{\varnothing }$.

If $X=\mathrm{\varnothing }$ then $F=\mathrm{\varnothing }$ 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 $X=W_0\cup \cdots \cup W_n$.

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 $F\mathrm{\ne }\mathrm{\varnothing }$, the case $F=\mathrm{\varnothing }$ being settled at step 2.1, and fix $x\in F$; then $x\in W_j$ for some $j\le n$, and $x\mathrm{\notin }X\setminus F$, so that $W_j\mathrm{\ne }X\setminus F$ 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 $\mathcal{V}:={\mathcal{V}}_0\cup \cdots \cup {\mathcal{V}}_n$ is a subfamily of $\mathcal{U}$; it is finite, a union of finitely many listable families being listed by concatenating their lists; and $K=K_0\cup \cdots \cup K_n\subseteq \bigcup \mathcal{V}$, since each $K_m$ lies inside $\bigcup {\mathcal{V}}_m\subseteq \bigcup \mathcal{V}$. So $\mathcal{V}$ is empty, in which case $K=\mathrm{\varnothing }$, or listable as $\{U_0,\dots ,U_p\}$ with $K\subseteq U_0\cup \cdots \cup U_p$; by [L2] the set $K$ is a compact subset of $X$, which is claim 2.

$F\subseteq V_0\cup \cdots \cup V_n$: given $y\in F$ there is $j\le n$ with $y\in W_j$, and $y\in F$ forces $W_j\mathrm{\ne }X\setminus F$, 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 $F=\mathrm{\varnothing }$ 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.