Under choice, the uncountable $\mathrm{\Delta }$-system lemma for finite sets

Partition the family $\mathcal{F}$ by finite cardinality. Some layer ${\mathcal{F}}_n=\{A\in \mathcal{F}:\mathrm{\mid }A\mathrm{\mid }=n\}$ is uncountable; otherwise [L1] would make their countable union $\mathcal{F}$ countable. It therefore suffices to prove the assertion by induction on the common size $n$.

The case $n=0$ is vacuous, since there is only one empty set.

Suppose the result holds for $(n-1)$-element sets and ${\mathcal{F}}_n$ is uncountable. If some point $x$ belongs to uncountably many members, apply the induction hypothesis to $\{A\setminus \{x\}:A\in {\mathcal{F}}_n,x\in A\}\mathrm{.}$ An uncountable $\mathrm{\Delta }$-subfamily with root $R$ then restores to one with root $R\cup \{x\}$.

It remains to suppose that ${\mathcal{F}}_n(x)=\{A\in {\mathcal{F}}_n:x\in A\}$ is at most countable for every $x$. Order the pairwise-disjoint subfamilies of ${\mathcal{F}}_n$ by inclusion. The union of a chain is again pairwise disjoint, so Zorn's lemma in [A1] gives a maximal such family $\mathcal{G}$.

If $\mathcal{G}$ were at most countable, then $M=\bigcup \mathcal{G}$ would be at most countable by [L1], because its members are finite. Maximality says every $A\in {\mathcal{F}}_n$ meets $M$, so ${\mathcal{F}}_n={\bigcup }_{x\in M}{\mathcal{F}}_n(x)\mathrm{.}$ The right side is a countable union of at most countable families and is at most countable by [L1], a contradiction. Hence $\mathcal{G}$ is uncountable.

The family $\mathcal{G}$ is pairwise disjoint, hence is an uncountable $\mathrm{\Delta }$-system with empty root. Together with step 1.3 this completes the induction and proves the lemma.