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

Partition the family $\mathcal F$ by finite cardinality. Some layer $\mathcal F_n=\{A\in\mathcal F:|A|=n\}$ 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 $\{A\setminus\{x\}:A\in\mathcal F_n,\ x\in A\}.$ An uncountable $\Delta$-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 $\mathcal F_n=\bigcup_{x\in M}\mathcal F_n(x).$ 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 $\Delta$-system with empty root. Together with step 1.3 this completes the induction and proves the lemma.