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The finite binary-function Cohen forcing $\mathrm{Fn}(\omega ,2,<\omega )$ is countable in $M$. It preserves ${\omega }_1$, and the countable levels, height, and normal extensions of the ground-model tree remain unchanged.

If the extension contained an uncountable tree antichain, apply the ground-model uncountable subset lemma for countable forcing to obtain an uncountable $X\in M$ contained in it. Incomparability in the fixed ground-model tree is absolute, so $M$ already regards $X$ as an uncountable tree antichain, contradicting that $T$ is Suslin in $M$. Similarly, a new cofinal branch has an uncountable ground-model subset. Comparability is absolute, so this would be an uncountable chain in the ground-model tree, again impossible.

Therefore countable forcing preserves Suslin trees, and in particular

An Aronszajn tree that is a union of countably many tree antichains. Equivalently, it admits a map into a countable set that is injective on each chain.

An Aronszajn tree with no uncountable tree antichain. A normal splitting Suslin tree yields a Suslin line by a lexicographic ordering followed by Dedekind completion.

A set of pairwise incomparable nodes in a set-theoretic tree. A maximal tree antichain has a comparable member for every node of the tree.