= Solution
Let $\langle D_\alpha:\alpha\in S\rangle$ 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
$$
\boxed{\diamondsuit_S\ \Longrightarrow\ \text{a nonspecial Aronszajn tree exists}.}
$$
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