(To lighten notations, let me simply write $\mathbb{R}$, instead of $\mathbb{R}^d$, for the set of Dedekind reals in what follows, since it is the only one that will appear. For the sake of notational consistency, I will also slightly change the notation for the decomposability of $\mathbb{R}$ to $\mathsf{DEC}(\mathbb{R})$ instead of $\mathit{dec}(\mathbb{R}^d)$.)
I claim that, assuming what is written in the question is correct, $\mathsf{DEC}(\mathbb{R})$ is equivalent to $\mathsf{WLPO}_{\mathbb{R}}$.
To see this, let me first remark that $\mathsf{DEC}(\mathbb{R})$ can be restated as the existence of a function $\chi\colon\mathbb{R}\to\{0,1\}$ such that $\exists a\in\mathbb{R}.\chi(a)=0$ and $\exists b\in\mathbb{R}.\chi(b)=1$. Replacing $\chi$ with $t \mapsto \chi(a+(b-a)t)$ we can assume $a=0$ and $b=1$, i.e., $\chi(0)=0$ and $\chi(1)=1$. Further replacing $\chi$ with $t \mapsto \chi(\inf(1,|t|))$, we can demand $\chi(t)=1$ when $|t|\geq 1$.
Proposition.$\mathsf{DEC}(\mathbb{R})$ together with (sequential) $\mathsf{WLPO}$ imply $\mathsf{WLPO}_{\mathbb{R}}$.
Proof. As remarked above, from $\mathsf{DEC}(\mathbb{R})$ we get a function $\chi\colon\mathbb{R}\to\{0,1\}$ such that $\chi(0)=0$ and $\chi(t)=1$ if $|t|\geq 1$.
Now let $c\in\mathbb{R}$. Define a binary sequence $p_n$ by $p_n = \chi(2^n c)$. If $c\mathrel{\#}0$, meaning $|c|>0$, then $\exists n.(p_n=1)$ (because there exists $n$ such that $2^{-n} \leq |c|$). Contrapositively, if $\forall n.(p_n=0)$ then $c=0$; but the converse is obvious. Thus, $(p_n)$ is the zero sequence iff $c=0$. Now sequential $\mathsf{WLPO}$ tells us that either $(p_n)$ is the zero sequence or it is not, so we get $(c=0) \lor \neg (c=0)$, which is $\mathsf{WLPO}_{\mathbb{R}}$. ∎
Now it is stated in the question (I did not try to prove this myself) that $\mathsf{DEC}(\mathbb{R})$ implies (sequential) $\mathsf{WLPO}$. So, together with the proposition above, this means that $\mathsf{DEC}(\mathbb{R})$ implies $\mathsf{WLPO}_{\mathbb{R}}$. The converse, of course, is obvious (since $\mathsf{WLPO}_{\mathbb{R}}$ asserts that $\mathbb{R}$ is decomposable as the disjoint union of $\{0\}$ and $\{x\in\mathbb{R} : \neg(x=0)\}$). So all of this proves that $\mathsf{DEC}(\mathbb{R})$ is equivalent to $\mathsf{WLPO}_{\mathbb{R}}$, as claimed at the start. (In particular, it implies $\mathsf{LLPO}_{\mathbb{R}}$, answering the titular question.)
