Re "Addition 2": Per Alexandersson's proof only regards the final step. At that point, every region is bounded by line segments (or rays/lines) and, due to the construction, every point at which multiple segments intersect, one of the segments continues all the way through - thus the arrangement determine only angles less than $\pi$, so the determined regions must be convex. The issue of boundaries is not related to the pigeonhole principle (which is a purely combinatorial statement) and the boundaries are explicitly handled by Joseph O'Rourke's addition.

One could add that we can consider points that are in between two collinear segments on a common extended segment to be in their own region (since having collinear segments decreases the number of "big" regions) - which , with your correction, then completely partitions the complement of the arrangement into exactly $n+1$ convex regions.

This is on M.SE as well. There are useful answers over on that page too.

This is cross-posted at MSE

I wonder if asking about $\sqrt{2}$ and $\sqrt{2}+\frac{1}3$ would be more tractable than $\sqrt{2}$ and $\sqrt{3}$ (I suspect that $\sqrt{2}$ being normal - or at least "random enough" - would at least tell us that that bitwise sum was irrational, though I lack proof)

@Jim It has no particular motivation, beyond the thought that a proof that $k<\text{lcm}(1,\ldots,n)$ for infinitely many $n$ would likely reveal something interesting about some "additional structure" allowing that to happen; in the other case, if $k=\text{lcm}(1,\ldots,n)$, it might be interesting to see why the exponent of a group is "irreducible"; I ask about the symmetric group in particular, because it has exactly one normal subgroup - so any approach applicable to it is likely to extend well to simple groups, but an approach could also leverage that $S_n$ is not itself simple.