The answer to the first question is yes. To sidestep the discussion about what it means for an unbounded set of reals to be measurable, I'll give an unambiguous example, namely that it's consistent for there to be a nonmeasurable subset of $[0, 1]$ which is a countable union of countable sets.

We work in ZF + "$\mathbb{R}$ is a countable union of countable sets." Let $\langle X_n: n<\omega \rangle$ be countable sets such that $\mathbb{R}=\bigcup_{n<\omega} X_n.$ Let $F_n$ be the countable subfield of $\mathbb{R}$ generated by $\bigcup_{i<n} X_i$ and $G_n = F_n \setminus F_{n-1}.$ In particular, $\mathbb{R}=\bigsqcup_{n<\omega} G_n.$

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} G_n \cap [0,1].$ If some $H_A$ is nonmeasurable, we're done since every set of reals is a countable union of countable sets. Otherwise, we will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 0.

Suppose $H_A$ has positive measure. By Lebesgue density theorem, there are arbitrarily small intervals $I$ with rational center such that $\lambda(H_A \cap I) > 0.9 \lambda(I).$ Since each $G_n$ is invariant under translations by rational numbers, this inequality holds for all sufficiently small intervals with rational center contained in $[0, 1].$ It would be impossible for this to also hold for $H_{\omega \setminus A},$ so $H_{\omega \setminus A}$ has measure 0.

Finally, notice that $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1].$