Balogh failure of countable shrinking

Suppose $(S_n{)}_{n<\omega }$ were separated sets covering $\kappa$. Replace each by its difference from the preceding finite union; the resulting disjoint sets still cover and retain separation by restriction of the witnesses. Let $f(\alpha )$ be the unique resulting piece index. Choose a witness $F_{\alpha }$ for each point and finite parameters $g(\alpha )\subseteq C$, $h(\alpha )\subseteq \kappa$ with $F_{\alpha }=F(\alpha ,g(\alpha ),h(\alpha ))$, using A1 and F2. Apply F1 to these typed functions. The resulting distinct $\alpha <\beta$ lie in the same piece because their $f$-values are equal. Their equations and $\beta \mathrm{\notin }h(\alpha )$ imply $\beta \in F_{\alpha }$ by F2, contradicting that piece's separation. Thus $\kappa$ cannot be such a countable union.

For $A\subseteq \kappa$ put $\mathrm{\Psi }(A)=A\cap \mathrm{\Phi }(A)$. If $\alpha \in A\setminus \mathrm{\Psi }(A)$, then $\alpha \mathrm{\notin }\mathrm{\Phi }(A)$, so F3 supplies an $F_{\alpha }\in {\mathcal{F}}_{\alpha }$ missing $A$. Choose these simultaneously using A1. For distinct $\alpha ,\beta$ in $A\setminus \mathrm{\Psi }(A)$, both are in $A$, giving $\beta \mathrm{\notin }{F}_{\alpha }$ and $\alpha \mathrm{\notin }{F}_{\beta }$. Hence this difference is separated. Empty differences use the empty witness family.

Let $C_0$ now denote a single closed subset of $U_n$, and define $A_0=\{\alpha :(\alpha ,0)\in C_0\}$ and $A_{j+1}=\mathrm{\Psi }(A_j)$ for $0\le j\le n$. These form a decreasing finite sequence. By induction, $A_j\times \{j\}\subseteq C_0$: the base is the definition, and at the next stage $A_{j+1}\subseteq \mathrm{\Phi }(A_j)$ puts its points on height $j+1$ in $\overline{A_j\times \{j\}}\subseteq C_0$ by F3 and closedness. Since $C_0\subseteq U_n$ has no point on height $n+1$, this proves $A_{n+1}=\mathrm{\varnothing }$. Consequently $A_0={\bigcup }_{j=0}^n(A_j\setminus A_{j+1})$. Each piece is separated by step 1.2, giving exactly the claimed finite decomposition. For $n=0$ there is one piece, $A_0\setminus A_1=A_0$.

If closed $C_n\subseteq U_n$ covered $L_0$, their bottom traces would cover $\kappa$. Apply step 2.1 to each $C_n$. Enumerating the pairs $(n,j)$ with $j\le n$ by successive finite rows turns those traces into a countable covering by separated sets, contrary to step 1.1. Therefore every such closed sequence misses some bottom point. If $X$ were countably paracompact, F4 applied to the increasing open cover from F3 would supply closed $C_n\subseteq U_n$ whose interiors cover $X$, hence whose sets cover $L_0$. This is impossible. All claimed failures follow. QED.