Here's a way to do it that also constructs a transitive irreflexive relation like bof's answer, but does so in a different way.
Define a relation $aSb$ as follows:
a----+----b for some x and y is the definition of aSb
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In other words, let $aSb$ abbreviate $\exists xy ( \lnot xRb \land aRb \land xRy \land yRb \land aRy \land aRx)$.
With this relation, we get strange behavior if self-edges are allowed. It holds that $aSb$ whenever $aRa$ and $aRb$ and $\lnot bRb$. This isn't actually a problem because we insist on the existence of a larger element for each element later, but ruling it out is more intuitive. So let's insist on $\forall a \mathop. \lnot aRa$.
Given the definition of $aSb$ above, it is a theorem that $\forall a \mathop. \lnot aSa$ holds because $aRx$ must be either true or false.
Transitivity is not free, we need to insist that $\forall x y z (xSy \land xSy \to xSz)$.
Antisymmetry is also not free, we must insist that $\forall a b (aSb \to \lnot bSa)$.
Totality is also not free, we must insist that $\forall a \exists b \mathop. aSb$.
If we put this all together we conceptually get a sprawling grid of incomplete $K_4$s expanding out in all directions.
$$ \forall a \mathop ( \lnot aRa) \land \forall x y z (xSy \land ySz \to xSz) \land \forall a b (aSb \to \lnot bSa) \land \forall a \exists b (aSb) $$
Our train of almost-$K_4$s can loop back on itself in weird ways, but insisting on transitivity, irreflexivity, and totality forces us to keep visiting new vertices at a conceptual level.