Nested intervals form a Suslin tree

Recursively choose $a_{\alpha }<b_{\alpha }$ in $L$ for every $\alpha <{\omega }_1$. At stage $\alpha$, the earlier endpoints form a countable set by F4. They cannot be dense in $L$, so some nonempty open interval $(c,d)$ misses all of them; density of $L$ supplies $c<a_{\alpha }<b_{\alpha }<d$. AC gives a choice operation on the nonempty sets of possible quadruples, and F5 then performs the recursion.

For $\xi <\alpha$, the connected interval $(c,d)$ selected at stage $\alpha$ avoids $a_{\xi },b_{\xi }$. Consequently either $[a_{\alpha },b_{\alpha }]\subset (a_{\xi },b_{\xi })$, or the two open intervals $(a_{\alpha },b_{\alpha })$ and $(a_{\xi },b_{\xi })$ are disjoint. Define $\xi \prec \alpha$ exactly in the first case. The relation is irreflexive and transitive. If $\xi ,\eta \prec \alpha$, their intervals both contain $[a_{\alpha },b_{\alpha }]$, so the disjoint alternative is impossible; whichever index is earlier is therefore $\prec$-below the other. Thus the predecessors of $\alpha$ are linearly ordered by the ordinal order and, being a subset of $\alpha$, are well ordered. Hence $({\omega }_1,\prec )$ is a tree.

There is no uncountable chain. Otherwise enumerate an uncountable chain increasingly as $({\alpha }_{\xi }{)}_{\xi <{\omega }_1}$. Successive nesting gives $a_{{\alpha }_{\xi }}<a_{{\alpha }_{\xi +1}}<b_{{\alpha }_{\xi +1}}<b_{{\alpha }_{\xi }}$, so the nonempty open intervals $(a_{{\alpha }_{\xi }},a_{{\alpha }_{\xi +1}})$ are pairwise disjoint. This contradicts ccc of $L$.

There is no uncountable antichain. For incomparable $\alpha ,\beta$, the dichotomy in step 2.1 makes $(a_{\alpha },b_{\alpha })$ and $(a_{\beta },b_{\beta })$ disjoint. An uncountable tree antichain would therefore give an uncountable family of pairwise disjoint nonempty open intervals in $L$, again contradicting ccc.

Each tree level is an antichain, hence countable by step 3.2. Every node has countable height because all its predecessors have smaller ordinal indices. If the tree height were a countable ordinal $\delta$, its carrier would be the union of the countably many countable levels indexed below $\delta$, hence countable by F4, contradicting that the carrier is ${\omega }_1$. Its height is therefore exactly ${\omega }_1$, not merely the length of the construction. A countable branch cannot be cofinal, because F4 makes the supremum of its countably many countable node-heights countable; an uncountable branch is excluded by step 3.1. Thus there is no cofinal branch. Steps 3.1-3.2 and F3 now show that the tree is Suslin. AC was used only for the recursion's simultaneous interval selections and the countable-union consequences recorded above.