Let $K$ be a number field and $p_1,p_2,\dots$ be a strictly increasing sequence of unramified primes. Then the set
$$\left\lbrace \mathfrak{p} : \mathfrak{p}\text{ divides }(p_j) \text{ for some } j\right\rbrace$$ is infinite. (Note $\mathfrak{p}$ is a prime ideal).
My attempt: Suppose the list of prime ideal divisors from $(p_1)$ to $(p_i)$ is $$P_i=\{\mathfrak{p}_1,\dots,\mathfrak{p}_r\}$$
Suppose the prime ideal decomposition of $(p_{i+1})$ is $\mathfrak{q}_1\dots\mathfrak{q}_s$. If $\mathfrak{q}_j\in P_i$ for all $1\leq j\leq s$, then $(p_{i+1})$ divides $$(p_1)\dots(p_i).$$ Hence, $p_{i+1}$ divides $$p_1\cdots p_i.$$ But $p_{i+1}$ is a prime that is larger than primes $p_1,\dots,p_i$. This is a contradiction. So $(p_{i+1})$ has a prime ideal divisor that is not yet listed.
This means that in each step, we add a new prime ideal in the list and so the lisy is infinite.
I'm just not sure that $p_{i+1}$ dividing $p_1\cdots p_i$ is a contradiction. I might be confusing primes with irreducibles or unknowningly assuming that $\mathcal{O}_K$ is a UFD.
The fact is true without unramified-ness or anything involving the order.
Take $1$ prime lying over each $p_i$. These have to be distinct because they lie over distinct primes, and there are infinitely many since there's one for each prime in the original infinite set. This is just a subset of your original set, and so that set too is infinite.