I feel ambiguity in the problem: whether $k$ should be identical for $m$ and $n$. My solution ends up settles the harder version. By the way, there are tons of other solutions.
Show that there exists a set $ A$ of positive integers with the following property: for any infinite set $ S$ of primes, there exist two positive integers $ m$ in $ A$ and $ n$ not in $ A$, each of which is a product of $ k$ distinct elements of $ S$ for some $ k \geq 2$.
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Proof:
For every $i\in\mathbb{N}$ associate the $i$-th prime $p_i$ with the binary representation of $i$. For every positive integer $k\ge2$, include in $A$ all products of $k$ distinct primes such that the sum of their respective $k-1$-th digits is odd.
Let $S$ contain two primes $p$ and $q$ that differ in their $k$-th digits. Since $S$ is infinite, pick $k$ other primes $q_1,q_2,\dots,q_{k}\in S$. By construction, exactly one of $$pq_1q_2\dots q_{k}$$ and $$qq_1q_2\dots q_{k}$$ is in $A$.
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