Example of an ordered set:
Let us define a set $C$.
Let $C$ is the set of circles of all radii. The circles are all lying on a plane. (The point circle is excluded.) If we take any two circles $c_1,c_2\in C$ and merge their centres, we will see one circle including another or the two circles are congruent. That is, only one of $c_1\subset c_2$, $c_2\subset c_1$, $c_2\approx c_1$ holds.
$\subset$ means includes, $\approx$ means congruent.
$\therefore$, Law of Trichotomy holds. — (i)
Now, if for $c_1,c_2,c_3\in C$, where $c_1\subset c_2$, $c_2\subset c_3$; then $c_1\subset c_3$.
Therefore, Law of Transitivity holds. — (ii)
$\therefore$, (i) and (ii) $\Rightarrow$ $C$ is an ordered set.
I've made this example. Is this example correct? Or is there any wrong?
It’s not a linear order on the set of all non-degenerate circles in the plane: if $c_1$ and $c_2$ are both circles of radius $1$, but they have different centres, then $c_1\approx c_2$, but $c_1\ne c_2$. If $\langle C,\subset\rangle$ were a (strict) linear order, this would be impossible: the trichotomy property actually requires exactly one of $c_1\subset c_2$, $c_2\subset c_1$, and $c_1=c_2$ to hold, with equality as one alternative.
If you let $C$ be the set of non-degerate circles in the plane that are centred at the origin (or at any other single point), then your $\approx$ actually becomes equality, and you do have a linear order. Alternatively, for each positive real number $r$ let $C_r$ be the set of all circles in the plane of radius $r$, let $\mathscr{C}=\{C_r:0<r\in\Bbb R\}$, and define $C_r\prec C_s$ if and only if $r<s$; then $\langle\mathscr{C},\prec\rangle$ is an ordered set.