= Solution
Fix $p$ and a name $\dot f$ such that $p$ forces a two-coloring of $[\lambda]^{<\omega}$. For each ground-model finite set $a\subseteq\lambda$, record its full <forcing decision pattern>
$$
F(a)(q)=\begin{cases}
0,&q\leq p\text{ and }q\Vdash\dot f(\check a)=0,\\
1,&q\leq p\text{ and }q\Vdash\dot f(\check a)=1,\\
2,&\text{otherwise}.
\end{cases}
$$
There are at most $3^{|\mathbb P|}<\lambda$ such patterns, since $\lambda$ is a <strong limit cardinal> and $|\mathbb P|<\lambda$. For finite $\mathbb P$ the number of patterns is finite, also below $\lambda$. Encode the patterns by an ordinal $\beta<\lambda$ and apply the assumed <finite-subset partition property> to obtain $H\subseteq\lambda$ of size $\lambda$ such that $F$ is constant on $[H]^n$ for every $n<\omega$.
For each $n$, choose one $a_n\in[H]^n$. Conditions below $p$ deciding $\dot f(a_n)$ are dense. Whenever one such condition decides the value, equality of patterns means it forces that same value at every $a\in[H]^n$. A <generic filter> containing $p$ meets this dense set for each $n$, so $f$ is constant on each $[H]^n$ in the extension. The constants may differ with $n$, exactly as required.
The forcing has size below the regular $\lambda$, hence satisfies the $\lambda$-<chain condition for forcing> and preserves its cardinality. The ground-model set $H$ consequently still has size $\lambda$. Since the argument works for every name and condition,
$$
\boxed{\Vdash_{\mathbb P}\lambda\rightarrow(\lambda)^{<\omega}_2.}
$$
This is <small forcing preservation of finite-subset partition properties>; no closure assumption on the forcing is required.
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