Let $\langle x\rangle$ denote the fractional part of a real number $x$ (i.e. $\langle x \rangle := x- \lfloor x\rfloor $, where $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$).
Let $\alpha \in \mathbb R$ be irrational and let $S:=\{n\in \mathbb Z: \langle n\alpha \rangle \in (0,1/4)\}$. The upper (and lower) density of $S$ is $1/4$; this is a consequence of Weyl's theorem on uniform distribution. Also, $S-S\subseteq \{n\in \mathbb Z: \langle n\alpha\rangle \in (3/4,1)\cup [0,1/4)\}$.
To see that $S-S$ does not contain an infinite arithmetic progression $\{a+bn:n\in \mathbb N\}$, note that $b\alpha$ is irrational if $b\in \mathbb Z\setminus \{0\}$, so the values $\langle (a+bn)\alpha \rangle$ are dense in $[0,1]$. So if $S-S$ contained an infinite AP, the values $\{\langle n\alpha \rangle:n\in S-S\}$ would be dense in $[0,1]$, but $\langle n\alpha\rangle \in (3/4,1)\cup [0,1/4)$ for $n\in S-S$.
This example $S$ is a Bohr neighborhood in $\mathbb Z$. Generally, if you want an example or counterexample of some structure in $S-S$, where $S$ has positive upper density, it's natural to look among Bohr neighborhoods: Følner ZBL0058.02302 proved that if $S$ has positive upper Banach density, then $S-S$ contains (up to upper Banach density 0) a Bohr neighborhood of $0$. Since every Bohr neighborhood $B$ of $0$ contains a set of the form $B'-B'$, where $B'$ is a Bohr neighborhood, $S-S$ itself is not too far from containing a difference set of a Bohr neighborhood.
Ruzsa's section Sumsets and structure in ZBL1221.11026 and Hegyvári and Ruzsa's article ZBL1333.05042 are both good references on the relationship between Bohr sets and difference sets.
The answer is: zero.
The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we will have a set in $\mathcal{U}$ with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.
Best Answer
Yes. As Martin says, this is often how the theorem is stated. It also follows immediately from the also common finitary form:
For all $\delta>0$ and $k\geq 1$, if $N$ is large enough depending on $\delta$ and $k$, $P$ is an arithmetic progression of length $N$, and $A\subseteq P$ has size $\lvert A\rvert\geq \delta N$, then $A$ contains a non-trivial arithmetic progression of length $k$.
(This is e.g. equivalent to the finitary form stated in the Wikipedia article where $P=\{1,\ldots,N\}$, since the property of containing an arithmetic progression is invariant under dilations and translations.)