Yes. Enumerate all arithmetic sets $A_1, A_2, \ldots$. Choose $n_k\in A_k$ with $|n_k|>2^k$. The set $\{n_1, n_2, \ldots\}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.

Actually you may construct such $S$ for each infinite sequence $A_1, A_2, \ldots$ of infinite sets: on $k$-th step choose $n_k\in A_k\cup S$ and $m_k\in A_k\setminus S$ so that $n_1, m_1, n_2, m_2, \ldots$ remain distinct.

Yes. Enumerate all arithmetic sets $A_1, A_2, \ldots$. Choose $n_k\in A_k$ with $|n_k|>2^k$. The set $\{n_1, n_2, \ldots\}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.

Actually you may construct such $S$ for each infinite sequence $A_1, A_2, \ldots$ of infinite sets: on $k$-th step choose $n_k\in A_k\cap S$ and $m_k\in A_k\setminus S$ so that $n_1, m_1, n_2, m_2, \ldots$ remain distinct.

Yes. Enumerate all arithmetic sets $A_1, A_2, \ldots$. Choose $n_k\in A_k$ with $|n_k|>2^k$. The set $\{n_1, n_2, \ldots\}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.

Yes. Enumerate all arithmetic sets $A_1, A_2, \ldots$. Choose $n_k\in A_k$ with $|n_k|>2^k$. The set $\{n_1, n_2, \ldots\}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.

Actually you may construct such $S$ for each infinite sequence $A_1, A_2, \ldots$ of infinite sets: on $k$-th step choose $n_k\in A_k\cup S$ and $m_k\in A_k\setminus S$ so that $n_1, m_1, n_2, m_2, \ldots$ remain distinct.