In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones

For claim 1 fix a compact $K\subseteq X$ and a point $x\in X\setminus K$, and put $\mathcal{V}:=\{V\in \mathcal{T}:U\cap V=\mathrm{\varnothing }\text{ for some }U\in \mathcal{T}\text{ with }x\in U\}$, a family cut out by a property of $V$ alone and not by any selection.

$K\subseteq \bigcup \mathcal{V}$: given $y\in K$ we have $y\mathrm{\ne }x$, since $x\mathrm{\notin }K$, so [A1] provides $U,V\in \mathcal{T}$ with $x\in U$, $y\in V$ and $U\cap V=\mathrm{\varnothing }$; that $V$ belongs to $\mathcal{V}$ and contains $y$.

If $K=\mathrm{\varnothing }$ then $U:=X$ and $V:=\mathrm{\varnothing }$ satisfy claim 1; otherwise [L2] applied to the family $\mathcal{V}$ gives $n\in \mathbb{N}$ and $V_0,\dots ,V_n\in \mathcal{V}$ with $K\subseteq V_0\cup \cdots \cup V_n$.

For each $j\le n$ the set $S_j:=\{U\in \mathcal{T}:x\in U\text{ and }U\cap V_j=\mathrm{\varnothing }\}$ is nonempty, because $V_j\in \mathcal{V}$; and $j\mapsto S_j$ is a function with domain the natural number $\sigma (n)$, so a choice function for its values supplies $U_0,\dots ,U_n\in \mathcal{T}$ with $x\in U_j$ and $U_j\cap V_j=\mathrm{\varnothing }$ for every $j\le n$.

Put $U:=U_0\cap \cdots \cap U_n$ and $V:=V_0\cup \cdots \cup V_n$; both are open by [L1], $x\in U$ because $x\in U_j$ for every $j$, $K\subseteq V$ by step 3.1, and $U\cap V=\mathrm{\varnothing }$ because a point of $U\cap V$ would lie in some $V_j$ and in $U\subseteq U_j$, contradicting $U_j\cap V_j=\mathrm{\varnothing }$. So claim 1 holds.

For claim 3 let $K\subseteq X$ be compact and put $G:=\bigcup \{W\in \mathcal{T}:W\cap K=\mathrm{\varnothing }\}$, which is open by [L1]. Every member of the union misses $K$, so $G\subseteq X\setminus K$; conversely for $x\in X\setminus K$ claim 1, proved at step 5.1, gives disjoint open $U\ni x$ and $V\supseteq K$, whence $U\cap K=\mathrm{\varnothing }$ and $x\in U\subseteq G$. So $G=X\setminus K$ is open, $K$ is closed, and claim 3 holds.

For claim 2 let $K,L\subseteq X$ be compact with $K\cap L=\mathrm{\varnothing }$, and put $\mathcal{W}:=\{W\in \mathcal{T}:V\cap W=\mathrm{\varnothing }\text{ for some }V\in \mathcal{T}\text{ with }K\subseteq V\}$, again cut out by a property. Then $L\subseteq \bigcup \mathcal{W}$: for $y\in L$ we have $y\mathrm{\notin }K$, so claim 1, proved at step 5.1, gives disjoint open $U\ni y$ and $V\supseteq K$, and that $U$ lies in $\mathcal{W}$ and contains $y$.

If $L=\mathrm{\varnothing }$ then $U:=\mathrm{\varnothing }$ and $V:=X$ satisfy claim 2; otherwise [L2] applied to $\mathcal{W}$ gives $m\in \mathbb{N}$ and $W_0,\dots ,W_m\in \mathcal{W}$ with $L\subseteq W_0\cup \cdots \cup W_m$.

For each $j\le m$ the set $T_j:=\{V\in \mathcal{T}:K\subseteq V\text{ and }V\cap W_j=\mathrm{\varnothing }\}$ is nonempty, because $W_j\in \mathcal{W}$; and $j\mapsto T_j$ is a function with domain the natural number $\sigma (m)$, so a choice function for its values supplies $V_0,\dots ,V_m\in \mathcal{T}$ with $K\subseteq V_j$ and $V_j\cap W_j=\mathrm{\varnothing }$ for every $j\le m$.

Put $U:=W_0\cup \cdots \cup W_m$ and $V:=V_0\cap \cdots \cap V_m$; both are open by [L1], $L\subseteq U$ by step 7.1, $K\subseteq V$ because $K\subseteq V_j$ for every $j$, and $U\cap V=\mathrm{\varnothing }$ because a point of $U\cap V$ would lie in some $W_j$ and in $V\subseteq V_j$, contradicting $V_j\cap W_j=\mathrm{\varnothing }$. So claim 2 holds.

For claim 4 assume $(X,\mathcal{T})$ is also compact: a compact subset of $X$ is closed by step 6.1, and a closed subset of $X$ is compact by [L4], so the two classes of subsets coincide; with claims 1, 2 and 3 settled at steps 5.1, 9.1 and 6.1 the theorem is proved.