Prove: R is an equivalence relation. In $ \mathbb{R^2}$ we consider the relation $(x,y)R(a,b)$ if and only if exists $n \in  \mathbb{Z} $ such that $n-1<y\leq n \ $ and $ n-1<b\leq n \  $. Prove: $ R$ is an equivalence relation.
$R$ is an equivalence relation if it is  reflexive, symmetric and  transitive .
Reflexive: $\forall  (x,y)\in \mathbb{R^2} \Rightarrow (x,y)R(x,y) \Rightarrow n-1<y\leq n \ \wedge n-1<y\leq n  $
Symmetric: $\forall  (x,y)\in \mathbb{R^2} \Rightarrow (x,y)R(a,b) \Rightarrow n-1<y\leq n \ \wedge n-1<b\leq n\Rightarrow n-1<b\leq n \ \wedge n-1<y\leq n\Rightarrow (a,b)R(x,y)  $
Transitive $\forall  (x,y)\in \mathbb{R^2}   $
\begin{split}
(x,y)R(a,b)& \| n-1<y\leq n \ \wedge n-1<b\leq n \\ (a,b)R(c,d)& \|n-1<b\leq n \ \wedge n-1<d\leq n  \\ &\Rightarrow n-1<y\leq n \ \wedge n-1<d\leq n\\&\Rightarrow\ (x,y)R(c,d)
\end{split}
Is correct my proof ?
 A: the last one should read
\begin{split}
(a,b)R(c,d)& \| n-1<b\leq n \ \wedge n-1<d\leq n \\
    &\Rightarrow n-1<y\leq n \ \wedge n-1<d\leq n\\
    &\Rightarrow\ (x,y)R(c,d)
\end{split}
(first evaluation is in $b$, not in $y$)
Also now will need to prove same $n$ for $y,b$ and $d$...
A: Your proof looks good to me, if they issues are fixed that others have pointed out. Using more words is good advice.
Another way to verify this is to make sure that the seeming equivalence relation generates well-defined equivalence classes that partition the underlying set. This particular relation appears to partition the plane into horizontal stripes: $\{(x,y)|n-1<y\leq n\}$ for each integer $n$. Thus, it appears to be an equivalence relation.
A: To expand on my comment, here's how I would adapt your proof.
Fix $(x,y),(a,b),(c,d)\in\mathbb{R}^2$.
For reflexivity, note that $\cup_{n\in\mathbb{Z}}(n-1,n]=\mathbb{R}$ and hence there exists $n\in\mathbb{Z}$ such that $n-1 < y \le n$. (Alternatively you can define $n:=\lceil y \rceil$ if you are familiar with the ceiling function.) Thus $(x,y)R(x,y)$.
For symmetry, suppose $(x,y)R(a,b)$. Then there exists $n\in\mathbb{Z}$ satisfying
$$
n-1 < y \le n \quad \text{and} \quad n-1 < b \le n,
$$
which by the symmetry of "and", is equivalent to 
$$
n-1 < b \le n \quad \text{and} \quad n-1 < y \le n.
$$
Consequently $(a,b)R(x,y)$.
For transitivity, suppose $(x,y)R(a,b)$ and $(a,b)R(c,d)$. This means there are
$n,m\in\mathbb{Z}$ such that
$$
n-1 < y \le n \quad \text{and} \quad n-1 < b \le n
$$
and
$$
m-1 < b \le m \quad \text{and} \quad m-1 < d \le m.
$$
Since $(n-1,n]$ and $(m-1,m]$ are disjoint whenever $n\ne m$ and $b\in(n-1,n]\cap(m-1,m]$, we infer $n=m$.
Hence $n-1=m-1 < d \le m = n$ so that we have
$$
n-1 < y \le n \quad \text{and} \quad n-1 < d \le n.
$$
Therefore $(x,y)R(c,d)$.
This completes the proof that $R$ is an equivalence relation.
