Let $S^1 = \{ (x,y) \in \mathbb R^2 : x^2 + y^2 = 1 \}$ and let $A = \{(0,-1), (0,1)\}$ and let $B = A \cup \{(1,0)\}$
A. Show that quotient spaces $S^1/A$ and $S^1/B$ are non-homeomorphic Hausdorff spaces.
B. Find closed set $X$ in space $S^1/A$ and continuous transformation $f: X \rightarrow D^2$ on boundary of disk $D^2 = \{ (x,y) \in \mathbb R^2 : x^2 +y^2 \le 1\}$ such that $(S^1/A) \cup_f D^2$ is homotopically equivalent with $S^1$
My approach
A. We know that the topological space $(X, T)$ is a Hausdorff space if for each $x_1 \neq x_2$ where $x_i \in X$ exists $U_j \in T$ such that $x_j \in U_j $ and $U_1 \cap U_2 = \emptyset$
I think that non-homeomorphism can be proved by the fact that $ S ^ 1 / A $ is divided by two points, and $ S ^ 1 / B $ by three, but I don't know how to present it formally or how to prove Hausdorff
B. I don't know how to look for such $ X $. May I ask you for help and how to do something like that?