One can actually show that dv($\kappa$) = $2^{<\kappa}$ outright. Suppose $D$ is a dense subset of $2^{\kappa}$ (or $2^{<\kappa}$, as preferred). I'll argue that $D$ must have size at least $2^{<\kappa}$, which is sufficient for the claim. If $r \in B(\kappa)$ is a bounded subset of $\kappa$, then (viewing $r$ as a characteristic function) let $\alpha_r$ be the least ordinal s.t. $\forall \beta \geq \alpha_r$, $r(\beta)=0$. We may think of $r$ as identifying an interval in $L(\kappa)$, namely the set of all points beginning with the sequence $r \upharpoonright \alpha_r$. Since $D$ is dense, there must be $x \in D$ that hits this interval, i.e. $x \upharpoonright \alpha_r = r \upharpoonright \alpha_r$. This shows that if we let $D'$ be the collection of all initial segments of elements of $D$, then $D' = 2^{<\kappa}$. But $|D'| \leq \kappa \cdot |D| = |D|$ which gives $2^{<\kappa} \leq |D|$.

One can actually show that dv($\kappa$) = $2^{<\kappa}$ outright. Suppose $D$ is any dense subset of $2^{\kappa}$ (one may take $D \subseteq B(\kappa)$, if preferred). I'll argue that $D$ must have size at least $2^{<\kappa}$, which is sufficient for the claim. For any sequence $r \in 2^{<\kappa}$, let $\alpha_r$ denote its length. One may think of $r$ as identifying the interval $I_r = \{x \in 2^{\kappa} \, : \, x \upharpoonright \alpha_r = r\}$. Since $D$ is dense in $2^{\kappa}$, there must be $x \in D$ that hits $I_r$, i.e. some $x$ beginning with the sequence $r$. This shows that if we let $D'$ be the collection of all initial segments of elements of $D$, then $D' = 2^{<\kappa}$. But $|D'| \leq \kappa \cdot |D| = |D|$ which gives $2^{<\kappa} \leq |D|$.

In the spirit of some of the comments, one can actually show that dv($\kappa$) = $2^{<\kappa}$ outright. Suppose $D$ is a dense subset of $2^{\kappa}$ (or $2^{<\kappa}$, as preferred). I'll argue that $D$ must have size at least $2^{<\kappa}$, which is sufficient for the claim. If $r \in B(\kappa)$ is a bounded subset of $\kappa$, then (viewing $r$ as a characteristic function) let $\alpha_r$ be the least ordinal s.t. $\forall \beta \geq \alpha_r$, $r(\beta)=0$. We may think of $r$ as identifying an interval in $L(\kappa)$, namely the set of all points beginning with the sequence $r \upharpoonright \alpha_r$. Since $D$ is dense, there must be $x \in D$ that hits this interval, i.e. $x \upharpoonright \alpha_r = r \upharpoonright \alpha_r$. This shows that if we let $D'$ be the collection of all initial segments of elements of $D$, then $D' = 2^{<\kappa}$. But $|D'| \leq \kappa \cdot |D| = |D|$ which gives $2^{<\kappa} \leq |D|$.

One can actually show that dv($\kappa$) = $2^{<\kappa}$ outright. Suppose $D$ is a dense subset of $2^{\kappa}$ (or $2^{<\kappa}$, as preferred). I'll argue that $D$ must have size at least $2^{<\kappa}$, which is sufficient for the claim. If $r \in B(\kappa)$ is a bounded subset of $\kappa$, then (viewing $r$ as a characteristic function) let $\alpha_r$ be the least ordinal s.t. $\forall \beta \geq \alpha_r$, $r(\beta)=0$. We may think of $r$ as identifying an interval in $L(\kappa)$, namely the set of all points beginning with the sequence $r \upharpoonright \alpha_r$. Since $D$ is dense, there must be $x \in D$ that hits this interval, i.e. $x \upharpoonright \alpha_r = r \upharpoonright \alpha_r$. This shows that if we let $D'$ be the collection of all initial segments of elements of $D$, then $D' = 2^{<\kappa}$. But $|D'| \leq \kappa \cdot |D| = |D|$ which gives $2^{<\kappa} \leq |D|$.