= Solution
The finite binary-function <Cohen forcing> $\operatorname{Fn}(\omega,2,{<}\omega)$ 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 \b[<countable forcing preserves Suslin trees>], and in particular
$$
\boxed{M[G]\models\text{“}T\text{ is a Suslin tree”}.}
$$
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