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Use the equivalent tree formulation of the Suslin hypothesis: there is no normal, well-pruned Suslin tree of height ${\omega }_1$, with countable levels and no uncountable chains or antichains. This is equivalent to the linear-order formulation that every complete dense order without endpoints satisfying the countable chain condition for a linear order is separable.

If such a tree $T$ existed, use its nodes as forcing conditions, with a higher extension stronger. It is CCC because its antichains are countable. For every $\alpha <{\omega }_1$, the set $D_{\alpha }=\{t:\mathrm{ht}(t)\ge \alpha \}$ is dense because the tree is well-pruned. The assertion ${\mathrm{MA}}_{{\aleph }_1}$ is precisely that a CCC forcing and a family of at most ${\aleph }_1$ dense sets admit a filter meeting all of them.

Apply it to these dense sets. A directed filter in a tree is a chain: two compatible nodes are comparable, since both lie among the well-ordered predecessors of a common extension. Meeting every $D_{\alpha }$ makes this chain cofinal, contrary to the defining absence of uncountable chains in a Suslin tree. Consequently

A Suslin line is a dense linear order without endpoints that is order-complete, has the countable chain condition for a linear order, and is not separable in its order topology. Thus every disjoint family of nonempty open intervals is countable, but there is no countable order-dense subset. The completeness condition says that every nonempty bounded-above subset has a supremum. These requirements distinguish a Suslin line from the real line, which has the countable order-dense subset $\mathbb{Q}$.

The diamond theorem in the constructible universe gives $L\models ♢$, and $L$ satisfies ZFC. Apply the preceding construction inside $L$ with $S={\omega }_1^L$. It produces a normal splitting Suslin tree.

For completeness, such a tree yields a Suslin line. Order its nodes lexicographically using the two successors at every split, treating a node itself as a position between its two successor subtrees. Each node is thus a cut point between a left and a right subtree. This gives a dense linear order without endpoints. Every nonempty interval contains a whole cone above some node: for comparable endpoints use the successor cone of the descendant endpoint directed toward the other endpoint; for incomparable endpoints use the right-successor cone of the lower endpoint. Disjoint intervals therefore supply pairwise incomparable cone roots, so the order has the countable chain condition for a linear order. A countable collection of nodes has bounded heights; a cone based above that bound contains none of them, so it is not an order-dense subset. Passing to the Dedekind completion using proper cuts, so that no endpoints are added, preserves density, the countable chain condition for a linear order, and nonseparability. For nonseparability, a countable dense set in the completion would give a countable dense set of original nodes by choosing one original node between each distinct pair of its points. This contradicts the preceding height-bound argument. The result is a Suslin line.

Thus the Suslin hypothesis fails in $L$. The constructible universe theorem is a theorem of ZFC, so this is a relative-consistency argument, without an additional assumption that a transitive model exists:

A complete dense linear order without endpoints with the countable chain condition for a linear order, but with no countable order-dense subset.