Let $$A = \{z \in \mathbb{C} : 1 \leqslant |z - (1+i)| \leqslant 2\}$$ and $$B = \{z \in A : |z - (1-i)| = 1\}$$. Then, $$B$$

We are given:

$$A = \{z \in \mathbb{C} : 1 \leqslant |z - (1+i)| \leqslant 2\}$$

$$B = \{z \in A : |z - (1-i)| = 1\}$$

Set $$A$$ is an annular region (ring) centered at $$(1, 1)$$ with inner radius 1 and outer radius 2.

Set $$B$$ consists of points that lie in $$A$$ and also on a circle centered at $$(1, -1)$$ with radius 1.

The circle $$|z - (1-i)| = 1$$ is centered at $$(1, -1)$$ with radius 1.

The distance between the centers $$(1, 1)$$ and $$(1, -1)$$ is:

$$ d = \sqrt{(1-1)^2 + (1-(-1))^2} = \sqrt{0 + 4} = 2 $$

For the circle $$|z - (1-i)| = 1$$ (center $$(1,-1)$$, radius 1), the distance from any point on this circle to $$(1,1)$$ ranges from $$d - 1 = 1$$ to $$d + 1 = 3$$.

For points in $$A$$, we need $$1 \leqslant |z - (1+i)| \leqslant 2$$.

Since points on the circle $$|z-(1-i)|=1$$ have $$|z-(1+i)|$$ ranging from 1 to 3, and $$A$$ requires $$|z-(1+i)| \in [1, 2]$$, the intersection is a continuous arc of the circle where $$1 \leqslant |z-(1+i)| \leqslant 2$$.

Since the range [1, 2] overlaps with [1, 3], there is a non-trivial arc of the circle that lies in $$A$$. This arc contains infinitely many points.

Therefore, $$B$$ is an infinite set.

The answer is Option D: is an infinite set.