This question was motivated by a statement in Simpson's Subsystems of Second Order Arithmetic (second edition), p. 168.
It is straightforward to verify (in $\mathsf{RCA}_0$ for instance) that $\leq_{KB}$ is a dense liner [sic] ordering with no left endpoint and with the empty sequence $\langle \rangle$ as its right endpoint.
(My emphasis.)
That it's a linear ordering with no left endpoint and $\langle \rangle$ as its right endpoint is indeed straightforward to verify. So is the Kleene/Brouwer ordering dense, too?
Let $Seq$ be the set of codes for finite sequences of natural numbers, and let $KB$ be the set of all pairs $(\sigma, \tau) \in Seq \times Seq$ such that either $\sigma \supseteq \tau$ or
$\exists{j < \min(lh(\sigma), lh(\tau))} \, [ \sigma(j) < \tau(j) \wedge \forall{i < j} (\sigma(i) = \tau(i)) ].$
Now consider the following: $\langle 0 \rangle <_{KB} \langle \rangle$, so if the order is dense then there must be some $\upsilon$ such that $\langle 0 \rangle <_{KB} \upsilon <_{KB} \langle \rangle$. Clearly it can't be the case that $\langle 0 \rangle \supset \upsilon \supset \langle \rangle$. And there is no point that $\upsilon$ can diverge from the empty sequence at. So it must be the case that $\upsilon \supset \langle \rangle$ and there exists some $j$ such that $\upsilon(j) < 0$. But since there is no natural number $n < 0$ this can't be the case either. So we have a counterexample to the density of the Kleene/Brouwer ordering.