But why would the appearance of a test sphere to someone far away make any difference, when we've already established that, to an observer right next to the test sphere, it would appear spherical, *not* distorted? To me, that local observation is a much better gauge of whether space is isotropic than the observation of someone far away, particularly when the light going to the faraway observer could be distorted by the variation in gravity in between.
Perhaps I introduced confusion when I used the word "anisotropy" to refer to the fact that there is more distance between two neighboring 2-spheres in the Schwarzschild exterior region than Euclidean geometry would lead one to expect based on the areas of the spheres. The only thing that that shows to be non-isotropic is the "Euclidean-ness" of the space. I've already described in detail how that "anisotropy" goes away as you descend through the non-vacuum "shell" region; if you want a "physical" explanation of how that works, I would say it's because the non-Euclideanness of the space is due to the mass of the shell being below you, as you descend through the shell, less and less of its mass is below you. "Below" here just means "at a smaller radius", or, if I were to be careful about describing everything in terms of direct observables, "below" means "lying on a 2-sphere with a smaller area than the one you are on".
It's also worth noting, I think, that even the "non-Euclideanness" of the space, as I described it, is observer-dependent; it assumes that the non-Euclideanness is being judged using 2-spheres which are at rest relative to the shell. Observers who are freely falling through the exterior vacuum region will *not* see this anisotropy in the space that is "at rest" relative to them; in other words, if a freely falling observer were to measure the areas of two adjacent 2-spheres that were falling with him, and measure the distance between the two 2-spheres, he would find the relationship between those measurements to be exactly Euclidean.