Furstenberg's proof defines a topology where a set is open if it is a union of sequences
S(a,b), where: S(a,b)={an+b:n∈ℤ}=aℤ+b.
However, then it says this is equivalent to saying that a set
$U$ is open iff $\forall x \in U \space \exists a \ne 0 \space s.t. S(a,x) \subseteq U$
I am not sure how this easily follows from the defintion.
On
Both just say that the set $\{S(a,b): a, b \in \mathbb{Z}, a \neq 0 \}$ forms a base for the topology in question.
And this collection fulfills the two conditions to be a base for a topology: the union of all $S(a,b)$ is $\mathbb{Z}$ (even $S(1,1) = \mathbb{Z}$ and for $n \in S(a,b) \cap S(c,d)$ we can find $S(a',b') $ such that $n \in S(a', b') \subseteq S(a,b) \cap S(c,d)$ (check this).
Defining such a collection and defining it as a base is quite usual practice.
There is no real obstacle.
Note that we can also write $S(a,b)=\{\,t\in\Bbb Z:a\mid t-b\,\}$ Assume $U$ is open, say $U=\bigcup_{(a,b)\in I}S(a,b)$. Then there exists $(a,b)\in I$ such that $x\in S(a,b)$ and of course $S(a,b)\subseteq U$. Now observe that $S(a,b)=S(a,x)$ if $x=\in S(a,b)$: We have $x=ma+b$ for some $n\in\Bbb Z$ and then $$t\in S(a,b)\iff a\mid t-b\iff a\mid t-x\iff t\in S(a,x).$$
On the other hand, assume that $U$ is a set such that for each $x\in U$, there exists $a=a(x)$ such that $S(a(x),x)\subseteq U$. Then clearly, $U=\bigcup_{(a,b)\in I}S(a,b)$, where $I=\{\,(a(x),x): x\in U\,\}$.