Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive

Theorem
Let $S$ be a set.

Let $\mathcal R \subseteq S \times S$ be a non-transitive relation on $S$.

Let $T \subseteq S$ be a subset of $S$.

Let $\mathcal R \restriction_T \ \subseteq T \times T$ be the restriction of $\mathcal R$ to $T$.

Then $\mathcal R \restriction_T$ is not necessarily a non-transitive relation on $T$.

Proof
Proof by Counterexample:

Let $S = \left\{{a, b}\right\}$.

Let $\mathcal R = \left\{{\left({a, b}\right), \left({b, a}\right), \left({b, b}\right)}\right\}$.

$\mathcal R$ is a non-transitive relation, as can be seen by definition.

Now let $T = \left\{{b}\right\}$.

Then $\mathcal R \restriction_T \ = \left\{{\left({b, b}\right)}\right\}$.

So:
 * $\forall x, y \in T: \left({x, y}\right) \in \mathcal R \restriction_T \land \left({y, z}\right) \in \mathcal R \restriction_T \implies \left({y, z}\right) \in \mathcal R \restriction_T$

as can be seen by setting $x = y = z = b$.

So $\mathcal R \restriction_T$ is a transitive relation on $T$.

That is, $\mathcal R \restriction_T$ is not a non-transitive relation on $T$.

Also see

 * Properties of Relation Not Preserved by Restriction‎ for other similar results.