This is more of a long comment than an answer.

The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\langle X_n: n<\omega \rangle$ be a partition of $[0, 1]$ into countable sets. Consider $X=\{x+m: n\le m<\omega, x \in X_n\}.$ It's not hard to show that $X$ has infinite outer measure and zero inner measure. Also, responding to a discussion in the comments, $\{\frac{1}{x}: x \in X\}$ is null since it can be written as a union of a countable set and a set of arbitrarily small outer measure. While this set satisfies the definition of measurability Emil provided in the comments, it violates almost everything we expect measurable sets to satisfy.

I think a better definition would be that $X \subset \mathbb{R}$ is measurable if for all $\epsilon>0,$ there is an open cover $C$ of $X$ such that for all bounded intervals $I,$ the restriction of $C$ to $I$ has measure less than $\lambda^*(X \cap I)+\epsilon,$ though I don't know if there's any literature studying this. Anyway, under this definition, your first question has an affirmative answer as demonstrated by the previous example, and the second question seems nontrivial.