Let $ a \neq 0,$ $b$ integers, show that $N_{a,b} := \{a + nb: n \in \mathbb{Z}\}$ is basis for some topology on $\mathbb{Z}.$ It is Hausdorff? Each $N_{a,b}$ is closed, why?
I am having trouble in how to show that each $N_{a,b}$ is closed... I saw in wikipedia that $N_{a,b}^c = \mathbb{Z} - \cup_{j=1}^{a-1}N_{a,b+j}.$ But, how can I see this?
How do I check that it is Hausrdoff or not?

First of all, there is an error in your question: denoting
$$N_{a,b}= \{ a+nb : n \in \Bbb{Z} \} = a+b \Bbb{Z}$$
you need $b \neq 0$ and not $a \neq 0$ (in fact, for $b=0$ this is just the singleton $\{ a \}$: this would generate the discrete topology on $\Bbb{Z}$).
You can see that $N_{a,b}$ is simply a coset of the subgroup $b \Bbb{Z}$. To check that these form a basis for a topology of $\Bbb{Z}$ (called the profinite topology of $\Bbb{Z}$) we have to check the usual axiom of basis of a topology (which I will not write down). Let
$$x \in (a+b\Bbb{Z}) \cap (a' + b'\Bbb{Z})$$
or, equivalently, let $x$ satisfy the system of congruences
$$x \equiv a \mod{b} \\ x \equiv a' \mod{b'}$$
Then one can easily see that $x+bb', x+2bb', x+3bb', \dots$ are solutions for this system; hence
$$x \in x + bb' \Bbb{Z} \subseteq (a+b\Bbb{Z}) \cap (a' + b'\Bbb{Z})$$
and this was exactly what you had to check.
Now, to see that all these basis elements are closed, I make an example. Fix $b=2$. Then, in practice, you have
$$2\Bbb{Z} , 1+ 2 \Bbb{Z}$$
and all of these form a partition of $\Bbb{Z}$. Since both are open, their complement is closed: but they are complements each other! So these two sets are closed.
You can do the same for
$$b\Bbb{Z} , 1+ b\Bbb{Z} , \dots , (b-1)+b\Bbb{Z}$$
these are pairwise disjoint open sets, forming a partition of $\Bbb{Z}$. Since the complement of each of these is a union of basic open sets (hence it is open), each of these is also closed.
