Proof equivalence relation checking my answer Let's define relation $\mathcal R$ in $\Bbb R$ in the followwing way:
$x \mathcal R y \iff (x=y \space \lor \space \exists k \in \Bbb Z (x,y \in (k,k+1)))$ 
Prove that $x \mathcal R y$ is equivalence relation.
My answer is as follows, please check correctness of my answer:
1) Reflexivity
$\forall x \in \Bbb R \space x \mathcal R x \iff (x=x \space \lor \space \exists k \in \Bbb Z (x \in (k,k+1)))$ 
2) Symmetry
$\forall x,y \in \Bbb R \space \space x \mathcal R y \Rightarrow y \mathcal R x \iff ((x=y \space \Rightarrow  y=x \space) \lor ( \space \exists k \in \Bbb Z (x,y \in (k,k+1)) \Rightarrow  \space \exists k \in \Bbb Z (y,x \in (k,k+1))) $ 
3) Transitivity
$\forall x,y,z \in \Bbb R \space \space x \mathcal R y \land y \mathcal R z \Rightarrow  x \mathcal R z \iff ((x=y \space \land  y=z \Rightarrow x=z) \space \lor  (\space \exists k \in \Bbb Z (x,y \in (k,k+1)) \land  \space \exists k \in \Bbb Z (y,z \in (k,k+1)) \Rightarrow \space \exists k \in \Bbb Z (x,z \in (k,k+1)) ) $ 
 A: Your all-symbols argument is way too dense to read.  And you have made a logical error by distributing the relation conditions across the or operation.
Instead, structure your argument in paragraphs like this:

We claim that $\mathcal{R}$ is symmetric.  Suppose that $x,y \in \mathbb{R}$ satisfy $x\mathcal{R}y$.  ... Therefore $y \mathcal{R} x$.  Since $x$ and $y$ were arbitrary, $\mathcal{R}$ is symmetric.

What goes in the ...?  If $x\mathcal{R}$y, then $x=y$ OR $x,y \in (k,k+1)$ for some $k$.  Proceed in cases.  
A: Two problems:
First, you have correctly worked out what it means to be reflexive, but you have not yet shown that that is the case. That is, you now need to show that it is indeed the case that
$$x=x \lor \exists k \in \mathbb{Z} (x \in (k,k+1))$$
Second, you have not correctly worked out what it means for $R$ to be symmetric and transitive. For example, symmetry becomes:
$$x=y \lor \exists k \in \mathbb{Z} ( x,y \in (k,k+1)) \Rightarrow y=x \lor \exists k \in \mathbb{Z} ( y,x \in (k,k+1))$$
and once again, you would need to show that that is actually the case. Same for Transitivity
