number of prime ideals in algebraic integers

There are infinitely many primes in $\mathbb{Z}_K$. You can prove it by showing that for every prime $p$ of $\mathbb{Z}$, there exists a prime ideal $P$ of $\mathbb{Z}_K$ such that $P \cap \mathbb{Z} = p\mathbb{Z}$. This is called the lying over theorem. There are infinitely many prime ideals of $\mathbb{Z}$, hence infinitely many prime ideals of $\mathbb{Z}_K$.

In the specific case of rings of integers, you can mimic Euclid's proof over $\mathbb Q$. The key point is that every ideal $\mathfrak a \subset \mathcal O_K$ factors (uniquely) as a product of prime ideals.

Suppose that we have a finite set of non-zero prime ideals $\mathfrak p_1,\ldots,\mathfrak p_n\subset\mathcal O_K$. Then $\mathfrak p_1\cdots\mathfrak p_n\cap\mathbb Z$ is a non-trivial ideal of $\mathbb Z$, so is equal to $(n)$ for some $n\in\mathbb Z\setminus\{0,1\}$. In particular, $n\in \mathfrak p_i$ for all $i$

Now consider the ideal $(n+1)\mathcal O_K$. If $\mathfrak p_i\mid (n+1)\mathcal O_K$ for any $i$, then $n, n+1\in \mathfrak p_i$ and hence $1\in \mathfrak p_i$, a contradiction.

In particular, $(n+1)\mathcal O_K$ is divisible by a prime ideal not in $\{\mathfrak p_1,\ldots,\mathfrak p_n\}$ and the result follows.

For a different proof mimicking Euler's proof for $\mathbb Q$, see this question.