I need to verificate or falsify the relation. In this previouse post (Showing $ x-y\in\mathbb{Q}$ is an equivalence relation?) they just showed that $x\in\mathbb{R}, x-x=0\in\mathbb{Q}$
Does this mean, that I only have to proof that $x\in\mathbb{R}, x-x=0$, which is element of $\mathbb{Z}$? Because I thought that I have to proof it somehow for $(x-y)$, like $(x-y)-(x-y)=0$
Your question and notation are not very clear. I will try to restate it in a way that makes more sense to me.
You write:
$\{(x,y)\in\{\mathbb{Z}\times\mathbb{Z}\} | x-y\in\mathbb{Z}\}$
I interpret $(x,y)\in\{\mathbb{Z}\times\mathbb{Z}\}$ to mean you are taking some $x\in\mathbb{Z}$ and relating it to some $ y\in\mathbb{Z}$.
The condition for this relation is: $x-y\in\mathbb{Z}$
In other words, for integers x and y, we have xRy ("x is related to y") precisely when x - y is an integer.
Now, you want to prove that this relation is reflexive, that is, xRx.
Since x is an integer, you need to check the condition: $x-x\in\mathbb{Z}$
Since for all integers x, it is true that: $x-x =0\in\mathbb{Z}$, then the relation is reflexive.