If it's any consolation, the idea is to give another one or two of these courses and then compile it all into a book.

(I've marked this as CW since it's really just a glorified comment. But the book is worth your effort in searching for these kind of propositions that you're asking about.)

In general, $\mu^{<\lambda^+}=\mu^\lambda$ whenever $\lambda$ is non-zero.

If you look at the notes on Axiomatic Set Theory on my website, Theorem 9.2 has a construction of an Aronszajn tree which may be closer to what you imagine here.

Have you looked at the notes on my site? The part about projections is kinda early because a lot of students pestered me about it, but it's there.

Joel, in this scenario the measure algebra is just the club filter and the non-stationary sets. $\cal P(\kappa)/\rm NS$ is not a measure algebra.

I suppose that the bigger issue is how to define Dedekind-infinite in the absence of LEM. We already have a dozen definitions in ZF (and half of which are "not Dedekind-finite" for various equivalent definitions of DF). Which one is "the right one" when LEM is, um, excluded.

I actually started writing that line of argument, but I had to leave my keyboard and forgot. I'm glad you wrote it out, since it's probably much better than anything I'd have written here.

@CalliopeRyan-Smith: In the case of the model you describe, note that there is (1) no reason to expect that the set $A$ of Cohen reals satisfies $|A|+|A|=|A|$, and (2) some reals will not have a canonical support where we can inject every set into $A$ or even into $[A]^\omega$, and even worse, some might not even have a canonical enumeration for their support, and there is no definable order on $[A]^\omega$ anyway.

@CalliopeRyan-Smith: Unfortunately, this isn't true. At least your hopes part, for example in most Solovay model-type results, we have that $[\Bbb R]^\omega$ cannot be linearly ordered, but $\sf SVC(\Bbb R)$ holds, since $V=L(\Bbb R)$. What you need is an injective seed which is linearly ordered. That's not so easy to find, and outright impossible in many cases.

@CalliopeRyan-Smith: It's not clear to me which is the "$\omega_1$ Cohen model". Note that $|X\times\omega|=|X|$ if and only if $|X|=|X|+|X|$, so we really just need that every set is linearly orderable along with that. Luckily, Sageev's model (which is really the only model we really know about where $|X|+|X|=|X|$ holds for every infinite set $X$ and choice fails) satisfies that every set can be linearly ordered. (This is Lemma 9.30 in his massive paper.)