Can we prove the existence of a non-measurable subset using an uncountable well-ordered subset without $\sf DC$?

Can we prove in $\sf ZF \ +$ “Every subset of $\mathbb R$ is Lebesgue measurable.” that $\aleph_1 \not \leq | \mathbb R|$?

I know that if we also assume $\sf DC$, the claim holds, but I don’t know if it can be done without using choice at all, thank you.

In this post, I show that the main consequences of countable choice regarding Lebesgue measure follow from (and are equivalent to) the assertion that the class of measurable sets is closed under countable union: https://mathoverflow.net/a/393162/109573

In particular, all of these properties of Lebesgue measure follow from the assertion $``$all sets of reals are measurable.$"$

A while ago, I went through Raisonnier's article and convinced myself that his arguments can all be carried out in this context. In particular, the Fubini argument that the null ideal is closed under well-ordered union can be carried out with an ad hoc verification of the relevant integration properties. I don't think there were any other subtleties that could cause a problem in the absence of choice.