@ZuhairAl-Johar Take $\alpha'$ to be the $\omega+1$ of any model of "ZF + PP fails at $V_{\omega+1}$" which is not an $\omega$-model.

Let $X$ be a Dedekind-finite infinite set of reals, inheriting the metric from $\mathbb{R}.$ This is a complete metric space since all Cauchy sequences are eventually constant. $X$ is not a closed set of reals so it is easy to find decreasing $E_n$ closed in $X$ with empty intersection.

The use of AC is essential. Any Dedekind-finite infinite set of reals is a counterexample.

It’s not open, WPP (and thus CB*) implies DC, see Higasikawa’s Partition Principles paper.

That still doesn't work: for two disjoint pieces $A$ and $B,$ the three sets $A,$ $B,$ and $A \cup B$ add to 0 in the vector space.

"In particular, no finite, nonempty, symmetric difference of these pieces is measurable." The pieces themselves are only a set of cardinality $\mathfrak{c}$ and there are many linear dependencies among the $2^{\mathfrak{c}}$ unions of them. That being said, you're right the dimension is $2^{\mathfrak{c}},$ since cardinality equals dimension in infinite vector spaces over a finite field.

The bijection is definable. I've posted an answer to the linked question.

This is provable in $Z,$ even in second-order arithmetic. The principle $ATR_0$ is equivalent to every uncountable closed set in a Polish space containing a perfect subset.

Re the last paragraph: this sounds like an approach to resolving the plausible conjecture. If $V \neq W$ implies that countable choice fails in some extension, that combined with your other answer would prove the conjecture.

"Almost well-orderable" is a nice notion! The Bernstein set result I told you about can be phrased as "if $\mathbb{R}$ is almost well-orderable, then there is a Bernstein set."