Under choice, every metrizable space has $w(X)=d(X)$

Suppose first that $\kappa$ is infinite. The family $\mathcal B=\{B(d,q):d\in D,\ q\in\mathbb Q,\ q>0\}$ has cardinality at most $\kappa\cdot\aleph_0=\kappa$ by [L2] and [L3]. It is a basis: if $x\in U$ with $U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$, choose $d\in D$ with $d(x,d)<\varepsilon/3$, and then by [L2] choose a positive rational $q$ with $d(x,d)<q<\varepsilon-d(x,d)$. Thus $x\in B(d,q)\subseteq B(x,\varepsilon)\subseteq U$. Hence $w(X)\le\kappa=d(X)$.

Suppose $\kappa$ is finite. If $D=\varnothing$, density forces $X=\varnothing$ and both raw invariants are $0$. Otherwise $X=D$: if $x\notin D$, the finitely many positive distances $d(x,a)$ for $a\in D$ have a positive minimum, and a smaller ball about $x$ misses $D$, contradicting density. A finite metric space is discrete, since at each point a ball smaller than all distances to the other finitely many points is a singleton. The singleton family is a basis of size $|X|$, and every basis of a discrete space must contain each singleton; also every dense set must contain every point. Therefore $w(X)=|X|=d(X)=\kappa$.

Step 1.1 handles infinite density and step 1.2 handles finite density; combining the resulting upper bound with [L1] gives $w(X)=d(X)$ in every case.