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At a correctly guessing limit stage of a normal tree construction, make every new level node extend a member of the guessed maximal tree antichain. It follows that the antichain has no later member. This produces a Suslin tree from the stationary diamond principle.

Let $⟨D_{\alpha }:\alpha \in S⟩$ witness the stationary diamond principle: for every $A\subseteq {\omega }_1$, the set of $\alpha \in S$ with $D_{\alpha }=A\cap \alpha$ is stationary. We construct a normal splitting Suslin tree; this will be a nonspecial Aronszajn tree.

Construct its levels by recursion. Start with one root, and give every node two immediate successors. At a countable limit $\alpha$, the constructed portion $T\upharpoonright \alpha$ is countable. Through each of its nodes choose a cofinal branch of that portion, and put one new node above each chosen branch at level $\alpha$. This keeps the level countable and gives every earlier node an extension. Branches are identified by their predecessor chains, so nodes at a limit level are uniquely determined by their predecessors.

At a limit $\alpha \in S$, decode $D_{\alpha }$ as a candidate tree antichain $B$ of $T\upharpoonright \alpha$. If it is maximal, choose the branches just described to meet $B$. This is possible: for any node, maximality supplies a comparable member of $B$, and normality of the already constructed portion extends the larger of those two nodes to a cofinal branch up to $\alpha$. Then every node at level $\alpha$, and every later node, lies above a member of $B$. This is antichain sealing by diamond.

Here is a precise way to handle the coding. Give the countable level $\xi$ node codes in $[\omega \xi ,\omega \xi +\omega )$. There is a club set of countable limit $\alpha$ with $\omega \alpha =\alpha$, and on this club the nodes below $\alpha$ have exactly the relevant codes below $\alpha$. Empty unused codes are ignored. Thus any subset of the entire tree has an ordinal code set to which the stationary diamond principle applies.

Let $B$ now be any maximal tree antichain of the completed tree. There is a club set of $\alpha$ such that $B\cap (T\upharpoonright \alpha )$ is maximal in $T\upharpoonright \alpha$. Indeed, choose a comparable member of $B$ for each node; closure under the heights of these witnesses gives that club. Intersect it with the coding club. Stationary correct guessing supplies an $\alpha \in S$ on this intersection at which $B$ is sealed. A member of $B$ at or above level $\alpha$ would extend a member of $B$ below $\alpha$, contradicting the tree antichain property. So $B$ is contained in the countable portion below $\alpha$. Every tree antichain extends to a maximal one, hence every tree antichain is countable.

There is no cofinal branch of length ${\omega }_1$. Otherwise, choosing at each successor level the other successor of its branch node would give an uncountable tree antichain. Thus the resulting tree is a Suslin tree. A special Aronszajn tree is a union of countably many tree antichains; here those would all be countable and could not cover the ${\aleph }_1$ nodes. Therefore

For a stationary $S\subseteq {\omega }_1$, there are cofinal ladders $A_{\delta }\subset \delta$ of order type $\omega$, indexed by limit ordinals in $S$, such that every uncountable $X\subseteq {\omega }_1$ contains one of these ladders. This predicts a contained ladder, whereas stationary diamond principle predicts a whole initial segment.