In a book on complex analysis, the authors prove:
Given finitely many (non-trivial) arithmetic progressions of natural numbers $$a_1, a_1+d_1, a_1+2d_1, \cdots $$ $$a_2, a_2+d_2, a_2+2d_2, \cdots, $$ $$a_k, a_k+d_k, a_k+2d_k, \cdots, $$ their totality (union) is never $\mathbb{N}$.
Given: any $k$ (non-trivial) arithmetic progressions.
Let $S$ denote the set of those numbers which are not in any of above $k$ arithmetic progressions.
If $S$ would have been finite, then we can form new finitely many arithmetic progressions which covers $S$, and combining them with given progressions, we will cover $\mathbb{N}$ by finitely many arithmetic progressions, a contradiction.
Thus $S$, the complement of all the given $k$ progressions, must be infinite.
Question Does $S$ contain infinitely many primes?
A non-trivial arithmetic progression means, the common difference is not $1$, i.e. it is not of the type $$a, a+1, a+2, \cdots $$
Lemma: for a number $s\in S$, we can show $s+d_1d_2d_3...d_k$ is also in $S$.
Proof: Suppose $s'=s+d_1d_2d_3...d_k$ is not in $S$, then $s'$ is in one of the arithmetic sequence, say $s'=a_i+md_i$ for some $i$. Then we can express $s=a_i+(m-d_1d_2...d_{i-1}d_{i+1}...d_k)d_i$ and contradicts $s\in S$. QED.
Applying our lemma, we can find an arithmetic sequence in $S$ and the result follows from Dirchlet Thereom.
EDIT: This actually requires there exist an $s$ coprime with $d_1d_2...d_k$.
For example, if we let the finite arithmetic sequences be $2+3k$ and $1+6k$ then there won't be infinitely many primes in $S$ as $3$ is the only one.