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Conditions are countable successor-length sequences $⟨A_{\xi }\subseteq \xi ⟩$, ordered by end extension. A countable fusion construction decides a named subset below a fresh limit index and writes that trace at the index inside any named club. The generic sequence satisfies the diamond principle.

The constructible universe satisfies the diamond principle. Together with antichain sealing by diamond, this supplies a Suslin tree in $L$.

Fix a diamond principle sequence $D_{\alpha }\subseteq \alpha$. Construct a normal splitting set-theoretic tree of height ${\omega }_1$ with countable levels. At successors give every node two successors. At a countable limit stage $\alpha$, the set-theoretic tree below $\alpha$ is countable. Choose countably many cofinal branches through it covering all its nodes, and put one node at level $\alpha$ above each distinct chosen branch. This preserves extension to all higher levels and tree with unique limits.

Arrange a coding of each level into the ordinal block $[\omega \alpha ,\omega (\alpha +1))$. On the club set of limit fixed points of $\alpha \mapsto \omega \alpha$, the nodes coded below $\alpha$ are exactly the nodes of height below $\alpha$. At a limit stage, if $D_{\alpha }$ codes a maximal tree antichain of the current set-theoretic tree below $\alpha$, require every chosen branch to meet it. This is possible: for any starting node $t$, maximality provides a comparable tree antichain member; if above $t$, first extend to it, and if below $t$, it has already been met. Then extend along a sequence of heights cofinal in $\alpha$. If the prediction is not a maximal tree antichain, use the ordinary covering branches. Thus every level is countable and the construction remains normal.

Here is the full chain-condition verification. Let $A$ be a maximal tree antichain in the final set-theoretic tree. For every node $t$, choose a witness $a_t\in A$ comparable with $t$. There is a club set of countable limit stages $\alpha$ closed under these witness choices: starting from any bound, repeatedly bound the heights of witnesses for all the countably many nodes below the current stage, and take the supremum after countably many steps. At such an $\alpha$, $A\cap T_{<\alpha }$ is already maximal in $T_{<\alpha }$.

View $A$ as a subset of ${\omega }_1$ through the coding. Diamond gives stationarily many stages with $D_{\alpha }=A\cap \alpha$. Choose one also in the witness-closure club set and the coding club set. The construction at that stage seals this very tree antichain: every node of level $\alpha$ extends one of its members below $\alpha$, and so does every node at a later level. No such node can itself belong to $A$, since it is comparable with an earlier member of $A$. HenceEvery tree antichain extends to a maximal one, so the set-theoretic tree has no uncountable tree antichain. Its normal splitting also excludes uncountable branches by part (ii). It is therefore a Suslin tree. By the standard Suslin-tree characterization of Suslin hypothesis, diamond implies failure of Suslin hypothesis. The decisive step is antichain sealing by diamond, with maximality below a correctly guessed club set stage verified explicitly.

Use diamond-sequence forcing. A condition is a sequenceordered by end extension. Its countable descending chains have lower bounds: take their union and, if their lengths approach a new limit, add an arbitrary subset at that last index. Hence it is countably closed and preserves ${\omega }_1$. The union of the generic filter supplies $⟨A_{\xi }:\xi <{\omega }_1⟩$.

To prove the diamond principle, let a condition force that $\dot{X}\subseteq {\omega }_1$ and that $\dot{C}$ is a club set. Below any such condition build $p_n$ and strictly increasing countable ordinals ${\delta }_n$ so that ${\delta }_n>{\delta }_{p_n}$, $p_{n+1}$ forces ${\delta }_n\in \dot{C}$, decides $\dot{X}\cap {\delta }_n$, and has length past ${\delta }_n$. Unboundedness supplies ${\delta }_n$; countable closure allows deciding all the bits below it. The construction can be carried out in $M$.

Let $\delta ={\sup }_n{\delta }_n={\sup }_n{\delta }_{p_n}$. The union of the conditions has entries exactly below $\delta$ and decides a ground-model set $x=\dot{X}\cap \delta$. Extend that union by setting $A_{\delta }=x$. This condition forces $\delta \in \dot{C}$ by closure, and $A_{\delta }=\dot{X}\cap \delta$. The conditions giving a correct guess inside any named club set are therefore dense. ThusNo ground-model Continuum hypothesis is needed; this forcing is allowed to collapse higher cardinals.