Is there a well known nontrivial counterexample to this claim?

Question

Suppose we have $A\subseteq\mathbb{N}$ with the property that if $B\subseteq\mathbb{N}$ and $B$ is finite, then $\exists a\in A\setminus\{ 1\}$, $\forall b\in B$, $\gcd(a,b)=1$.

Are there any well known sets that are counterexamples to the claim that $A$ contains an infinite number of prime numbers?

The claim is certainly wrong, but the only counterexamples I know of are sets and constructions of the form $A=\mathbb{N}\setminus\mathbb{P}$ where $\mathbb{P}$ is the set of prime numbers. (To show this $A$ works, choose $a=p_B^2$ where $p_B$ is the first prime greater than $\max\{ B\}$). Is there a known counterexample that doesn't trivially discard primes in its construction?

The set of Fermat numbers would be a counterexample if it was proven it contained a finite number of primes.

Answer 1

$$A=\{ n^2 | n \in \mathbb Z \} $$ works.

If $B=\{ b_1,.., b_n \}$ is any finite set then set $a=(b_1....b_n+1)^2$.

Same way the set of any fixed power of integers works.

Answer 2

If you want $A$ to have that property you want, and you don't want $A$ to contain any prime numbers, why not take all the prime numbers $p_1, p_2, ...$ and let $$A = \{p_1p_2, p_2p_3, p_3p_4 ... \}$$