 Solution

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