The answer to your first question is yes, and the answer to your second question is no, under any of the multiple definitions of "measurable" in choiceless contexts.

We will prove a theorem relating various measure-theoretic consequences of countable choice.

(ZF) The following are equivalent. Note that (1)-(6) are about the algebra of subsets of $[0,1]$ which satisfy $\lambda^*(X)+\lambda^*([0,1] \setminus X)=1$ while (7) refers to the algebra of subsets of $\mathbb{R}$ which satisfy $\lambda^*([-n, n] \cap X) + \lambda^*([-n, n] \setminus X) = 2n$ for all $n.$

- Lebesgue measure is $\sigma$-additive.
- A countable union of measurable sets is measurable.
- A countable union of null sets is measurable.
- A countable union of null sets is null.
- Every null set is contained in a null $G_{\delta}$ set.
- For every measurable set $X,$ there is an $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset X \subset B$ and $B \setminus A$ is null.
- Any of the above but for measurable subsets of $\mathbb{R}.$

Proof:

(1) $\rightarrow$ (2) Clear.

(2) $\rightarrow$ (3) Clear.

(3) $\rightarrow$ (4) Suppose towards contradiction $X_i$ are null sets with $\lambda(\bigcup_{i<\omega} X_i)>0.$ Let $Y_i = \{x+\frac{m}{2^i}: m < 2^i, \exists j < i (x \in X_j)\}$ (note that addition is mod 1) and $Z_i = Y_i \setminus Y_{i-1}.$ In particular, the $Z_i$ are disjoint null sets whose union has positive measure and is closed under translation by dyadic rationals.

For $A \subset \omega,$ let $H_A=\bigcup_{n \in A} Z_n.$ By (3), each $H_A$ is measurable. We will show that for every $A \subset \omega,$ either $H_A$ or $H_{\omega \setminus A}$ has measure 1.

Suppose $H_A$ has positive measure (otherwise $H_{\omega \setminus A}$ has positive measure). Fix $\epsilon>0.$ By Lebesgue density theorem, there is some interval $I$ of length $\frac{1}{2^n}$ such that $\lambda(H_A \cap I) > \frac{1-\epsilon}{2^n}.$ Clearly $H_{A \setminus (n+1)}$ also satisfies this inequality, and is furthermore closed under translation by $\frac{1}{2^n}.$ Thus, $\lambda(H_A) = \lambda(H_{A \setminus (n+1)}) > 1-\epsilon,$ so $\lambda(H_A)=1.$

Since $H_{\omega}$ has measure 1, we see that $[0,1]$ is a countable union of null sets. By (3), every subset of $[0,1]$ is measurable. However, $\{A \subset \omega: H_A \text{ is measure 1} \}$ is a non-principal ultrafilter, so there is a nonmeasurable subset of $[0,1],$ contradiction.

(4) $\rightarrow$ (1) Let $X_i$ be measurable sets. Let $U_i$ enumerate the basic open sets. Define $S_n \subset \omega$ recursively by having $i \in S_n$ iff there is $j$ such that $\lambda(U_i \cap X_j \setminus \bigcup_{k<i, k \in S_n} U_k) > \frac{n}{n+1} \lambda(U_i).$ Let $V_n = \bigcup_{i \in S_n} U_i.$ Then $V:=\bigcap_{n < \omega} V_n$ is a $G_{\delta}$ set such that $\lambda(V)=\sup_{n<\omega} \lambda(\bigcup_{i < n} X_i)$ and $V \triangle \bigcup_{i<\omega} X_i$ is null.

(4) $\rightarrow$ (5) Let $X$ be null. By (4) we can assume $X$ is closed under translation by rational numbers. Let $U$ be an open cover of $X$ of measure less than $\frac{1}{2}.$ We can canonically cover $X \cap [0, \frac{1}{n}]$ with an open set of measure $\frac{1}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m}{n}, \frac{m+1}{n}])<\frac{1}{2n}.$ We can thus recursively construct open covers of $X$ of measure less than $\frac{1}{2^n}.$

(5) $\rightarrow$ (6) Similarly to in (4) $\rightarrow$ (1), there is a $G_{\delta}$ set $B_1$ with null symmetric difference from $X.$ Let $B_2$ be a null $G_{\delta}$ set containing $X \setminus B_1.$ Then $B:=B_1 \cup B_2$ is a $G_{\delta}$ set with $X \subset B$ and $B \setminus X$ null. We can similarly construct such a $B'$ for $[0, 1] \setminus X$ and set $A = [0, 1] \setminus B'.$

(6) $\rightarrow$ (4) Let $X_i$ be null sets. Let $X=\bigcup_{i<\omega} X_i.$ Consider $Y = \{2^{-i-1}(1+x): x \in X_i\}.$ It is easy to see $Y$ is null, and thus contained in some null $G_{\delta}$ set $Y'.$ Let $U_n$ be a sequence of open sets covering $Y,$ each satisfying $\lambda(U_n)<\frac{1}{n}.$

Fix $\epsilon>0$ and $i<\omega.$ Let $n$ be least such that $\frac{1}{2^n}<\frac{\epsilon}{4^{i+1}}.$ Then $X_i$ is covered by a translation of $U_n$ scaled up by $2^{i+1},$ which has measure less than $\frac{\epsilon}{2^{i+1}}.$ Applying this construction to all $i,$ we get a cover of $X$ of measure less than $\epsilon.$ Thus, $X$ is null.

(7) Finally, it is routine to verify that assertions (1)-(6) collectively prove their generalizations to $\mathbb{R}.$ E.g., one could verify $\frac{1}{x}$ on $(0, \infty)$ sends null sets to null sets and measurable sets to measurable sets
using the fact that it's Lipschitz on each $[2^{-n}, 2^n]$ and that $\frac{1}{x}$ sends $G_{\delta}$ null sets to $G_{\delta}$ null sets. $\square$

Thus, $\text{M}_{\omega}$ fails in any model of ZF + "$\mathbb{R}$ is a countable union of countable sets" since this theory negates (1), providing an affirmative answer to question 1. As for question 2, if $\neg \text{BNM}$ holds, then we immediately have (2), so every subset of $\mathbb{R}$ is measurable. In particular, all interpretations of "all sets are measurable" are equivalent.

definition) ... A set $A$ is measurable if and only if $A \cap [a,b]$ is measurable for every $a<b$. Are you claiming this is cannot be proved in ZF? Perhaps improve the question by including the definition of "measurable". $\endgroup$shouldactually work. I was originally thinking about arbitrary test sets $Y$. So perhaps this part of the definition is a crucial point that needs to be clarified. $\endgroup$23more comments