Here is an infinite set of primes, for which I think I can prove zero density:

Let $P$ be a set of primes {$p_1,\ldots,p_n,\ldots$} recursively defined such that $p_n\equiv 1 \bmod p_i$ for all $ i \lt n$.

1. The set $P$ is infinite by Dirichlet's theorem in arithmetic progressions, because if $p_1,\ldots, p_n$ are the first $n$ primes in $P$, then there is always a prime $p \equiv 1 \bmod p_1\cdots p_n$, and therefore $p\equiv 1 \bmod p_i$ for all $i=1,\ldots, n$.

2. The density of $P$ must be zero, because, if we fix $N>0$

$$\limsup_{x\to\infty}\frac{|\text{$p\leq x$ such that $p\in P$}|}{|p\leq x|} \leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N| + |\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|}$$ $$\leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N|}{|p\leq x|} + \frac{|\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|} = 0 + \frac{1}{\varphi(p_1\cdots p_N)} \to 0 \text{ as $N\to \infty$.}$$ Hence $\delta(P)=\lim_{x\to\infty}\frac{|p\leq x \text{ such that } p\in P|}{|p\leq x|}$ exists and it is zero.

Here is a simple example of how one can use Dirichlet's theorem on primes in arithmetic progressions to find infinite families of primes with zero density:

Let $P$ be a set of primes {$p_1,\ldots,p_n,\ldots$} recursively defined such that $p_n\equiv 1 \bmod p_i$ for all $ i \lt n$.

1. The set $P$ is infinite by Dirichlet's theorem in arithmetic progressions, because if $p_1,\ldots, p_n$ are the first $n$ primes in $P$, then there is always a prime $p \equiv 1 \bmod p_1\cdots p_n$, and therefore $p\equiv 1 \bmod p_i$ for all $i=1,\ldots, n$.

2. The density of $P$ must be zero, because, if we fix $N>0$

$$\limsup_{x\to\infty}\frac{|\text{$p\leq x$ such that $p\in P$}|}{|p\leq x|} \leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N| + |\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|}$$ $$\leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N|}{|p\leq x|} + \frac{|\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|} = 0 + \frac{1}{\varphi(p_1\cdots p_N)} \to 0 \text{ as $N\to \infty$.}$$ Hence $\delta(P)=\lim_{x\to\infty}\frac{|p\leq x \text{ such that } p\in P|}{|p\leq x|}$ exists and it is zero.

Here is an infinite set of primes, for which I think I can prove zero density:

Let $P$ be the set of all primes $p$ such that $p\equiv 1 \bmod q$ for all primes $q \lt p$.

1. The set $P$ is infinite by Dirichlet's theorem in arithmetic progressions, because if $p_1,\ldots, p_n$ are the first $n$ primes, then there is always a prime $p \equiv 1 \bmod p_1\cdots p_n$, and therefore $p\equiv 1 \bmod p_i$ for all $i=1,\ldots, n$.

2. The density of $P$ must be zero, because, if we fix $N>0$

$$\limsup_{x\to\infty}\frac{|\text{$p\leq x$ such that $p\in P$}|}{|p\leq x|} \leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N| + |\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|}$$ $$\leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N|}{|p\leq x|} + \frac{|\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|} = 0 + \frac{1}{\varphi(p_1\cdots p_N)} \to 0 \text{ as $N\to \infty$.}$$ Hence $\delta(P)=\lim_{x\to\infty}\frac{|p\leq x \text{ such that } p\in P|}{|p\leq x|}$ exists and it is zero.

Here is an infinite set of primes, for which I think I can prove zero density:

Let $P$ be a set of primes {$p_1,\ldots,p_n,\ldots$} recursively defined such that $p_n\equiv 1 \bmod p_i$ for all $ i \lt n$.

1. The set $P$ is infinite by Dirichlet's theorem in arithmetic progressions, because if $p_1,\ldots, p_n$ are the first $n$ primes in $P$, then there is always a prime $p \equiv 1 \bmod p_1\cdots p_n$, and therefore $p\equiv 1 \bmod p_i$ for all $i=1,\ldots, n$.

2. The density of $P$ must be zero, because, if we fix $N>0$

$$\limsup_{x\to\infty}\frac{|\text{$p\leq x$ such that $p\in P$}|}{|p\leq x|} \leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N| + |\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|}$$ $$\leq \lim_{x\to\infty}\frac{|p\leq p_1\cdots p_N|}{|p\leq x|} + \frac{|\text{$p$ such that $p\equiv 1 \bmod p_1\cdots p_N$}|}{|p\leq x|} = 0 + \frac{1}{\varphi(p_1\cdots p_N)} \to 0 \text{ as $N\to \infty$.}$$ Hence $\delta(P)=\lim_{x\to\infty}\frac{|p\leq x \text{ such that } p\in P|}{|p\leq x|}$ exists and it is zero.