To gain more insight on the non measurability of Vitali sets, I have been considering the following generalization.
$\color{#FF5200}{\bf{Definition\ 1.}}$ Given an additive subgroup $(G,+)\leq(\mathbb{R},+)$, we say that $E\subseteq\mathbb{R}$ is a choice of the quotient $\mathbb{R}/G$ if $\forall x\in\mathbb{R}\ \exists!e_x\in E:x-e_x\in G$. That is, if $+:G\times E\to\mathbb{R}$ is bijective. Vitali sets are choices of the quotient $\mathbb{R}/\mathbb{Q}$. We denote the set of quotient choices of $\mathbb{R}/G$ as $$\text{Ch}(\mathbb{R}/G):=\{E\subseteq\mathbb{R}\ \vert\ \forall x\in\mathbb{R}\ \exists!e_x\in E:x-e_x\in G\}$$
We know that $(\mathbb{Q},+)\leq(\mathbb{R},+)$ is such that no element of $\text{Ch}(\mathbb{R}/\mathbb{Q})$ is measurable. But note that not every additive subgroup of $\mathbb{R}$ satisfies this. For instance, $\text{Ch}(\mathbb{R}/\{0\})=\{\mathbb{R}\}$ and $\text{Ch}(\mathbb{R}/\mathbb{R})=\{\{x\}:x\in\mathbb{R}\}$ are families of measurable sets. With this in mind, I have been trying to characterize the subgroups $(G,+)\leq(\mathbb{R},+)$ that don't admit measurable choices of $\mathbb{R}/G$. Sadly, to no avail. However, I've found a necessary condition and a sufficient one.
- Necessary. If $\text{Ch}(\mathbb{R}/G)$ has no measurable sets, then $G\subseteq\mathbb{R}$ is dense and $\mathbb{R}/G$ is uncountable.
- Sufficient. If $G$ is dense and countable, then $\text{Ch}(\mathbb{R}/G)$ has no measurable sets.
The necessary condition is due to the fact that non dense subgroups of $(\mathbb{R},+)$ are of the form $a\mathbb{Z}$ so $[0,|a|)\in\text{Ch}(\mathbb{R}/a\mathbb{Z})$ which is measurable. And, if $\mathbb{R}/G$ were countable, then $\text{Ch}(\mathbb{R}/G)$ would only contain countable sets and, thus, measurable ones. For the sufficient condition, if $G$ is dense and countable, it can be proven analogously to the case of $\mathbb{Q}$ (Vitali sets) that $\text{Ch}(\mathbb{R}/G)$ can't have any measurable set. So, in $\sf ZFC+CH$ ($\sf CH$ to ensure the existence of Sierpiński sets),
$\bf{Q_1}\hspace{-2pt}:$ Can $(G,+)\leq(\mathbb{R},+)$ be uncountable and such that $\text{Ch}(\mathbb{R}/G)$ has no measurable sets?
$\bf{Q_2}\hspace{-2pt}:$ $G$ and $\mathbb{R}/G$ being both uncountable implies that $\text{Ch}(\mathbb{R}/G)$ has no measurable sets?
If the answer of $\bf{Q_1}$ were negative or the answer of $\bf{Q_2}$ were positive, we would then have (resp.) $$(G,+)\text{ admits no measurable choices of }\mathbb{R}/G\iff G\text{ is dense and countable}$$ $$(G,+)\text{ admits no measurable choices of }\mathbb{R}/G\iff G\text{ is dense and }\mathbb{R}/G\text{ is uncountable}$$
$\color{#FF8082}{\bf{A\ particular\ case.}}$ Consider $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Then, by the axiom of choice, we can find a Hamel basis $\mathcal{B}\subseteq\mathbb{R}$. If we partition it as $\mathcal{B}=\mathcal{B}_1\dot\cup\mathcal{B}_2$ so that $|\mathcal{B}_1|=|\mathcal{B}_2|=2^{\aleph_0}$, then $G:=\langle\mathcal{B}_1\rangle_\mathbb{Q}$ is an uncountable subgroup of $\mathbb{R}$ with uncountable index (i.e. $\mathbb{R}/G$ uncountable) as $\mathbb{R}/G\cong\langle\mathcal{B}_2\rangle_\mathbb{Q}$. Suppose that $\mathbb{Q}\cap\mathcal{B}_1\neq\emptyset$ so that $\mathbb{Q}\leq G$. Now I will show that every measurable set of $\text{Ch}(\mathbb{R}/G)$ has measure zero. If $E\in\text{Ch}(\mathbb{R}/G)$, then $+:G\times E\to\mathbb{R}$ is bijective so $+:\mathbb{Q}\times E\to\mathbb{R}$ is injective. Now, by transfinite induction (which is a theorem of $\sf ZFC$), we can extend $E\subseteq E^\#$ so that $+:\mathbb{Q}\times E^\#\to\mathbb{R}$ is bijective. But this means that $E$ is contained on a Vitali set $E^\#\in\text{Ch}(\mathbb{R}/\mathbb{Q})$. Hence, if $E$ were measurable, it would necessarily have measure zero. Also, since $E$ is uncountable, the existence of a measurable choice $E$ of $\mathbb{R}/G$ would imply the existence of a Vitali set that contains an uncountable set with measure zero and, thus, a Vitali set that won't be Sierpiński (which doesn't say anything).
Can it? Yes, it is consistent, at the very least, that the answer can be positive.
Assume that every set of reals of size strictly less than $2^{\aleph_0}$ is null, for example, assuming Martin's Axiom holds will provide us with this situation.
Now fix a Hamel basis of $\Bbb R$ over $\Bbb Q$, and consider $G$ to be the subspace generated by any co-"small uncountable subset" of the Hamel basis. It is easy to see that the dimension of the quotient is less than the continuum, so any choice of representatives will be null.
The real question you'd want to ask, then, is whether or not this is provable in $\sf ZFC$, and can we change "uncountable" to "size continuum".