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Let $x $ ~ $ y $ $:\iff$$x\leq y\lor y\leq x$

and $x,y\in M$ and $M$ is an ordered set.

I think this does define an equivalence relation with only one equivalence class.

Proof:

Reflexive:$x$~$x$
$x\leq x$

Symmetric:$x$~$y$ $\implies$$y$~$x$

If $x\leq y$ $\implies$ $y\geq x$

If $x\geq y$ $\implies$ $y\leq x$

Transitive:$x$~$y$ $\land$ $y$~$z$ $\implies$$x$~$z$

$x\leq y\lor y\leq x$ $\land$ $y\leq z\lor y\leq z$ $\implies$ $x\leq z\lor z\leq x$

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    Any element of natural numbers could be chosen as equivalence class represantative ? – 2017-02-15
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    If $x\sim y$ and $y\sim z$, then you coud have a case where $x\le y$ and $z\le y$, so in that case how do you know $x\sim z$? I don't think this is transitive. – 2017-02-15
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    How could i find a counterexample to transitivity ? – 2017-02-15
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    Try something with set inclusion. You need partial order to get a counterexample – 2017-02-15

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No. Your proof of transitivity is wrong (even after you fix the typo!). For a simple counterexample, take the power set of $\{a,b\}$, ordered by inclusion. Then $\{a\}\sim\{a,b\}$ and $\{b\}\sim\{a,b\}$, but $\{a\}\not\sim\{b\}$