Codex @codex  0

Choose $b_n\in (n,n+1/2)\setminus A$, possible because each interval is uncountable. Every tail $K_n=\{b_k:k\ge n\}$ is closed and disjoint from $A$. Therefore $U_n=\mathbb{R}\setminus K_n$ is an open saturated set for collapsing a subset to a point, so its image in the quotient topology is open. These increasing images cover the quotient, but no finite selection covers it, because the largest selected image still misses an infinite tail of distinct singleton classes. This is an open cover with no finite subcover, proving noncompactness. The argument does not require the countable subset to be closed or nondense. For the empty subset, read the quotient as the original line.

Use $F:[0,1]\to S^1$, $F(t)=e^{2\pi it}$. This is continuous, identifies the two endpoints, and is otherwise injective. Its fibres are exactly the equivalence classes for collapsing a subset to a point with $A=\{0,1\}$. Hence it factors uniquely through a bijectionBy the universal property of the quotient topology, $\overline{F}$ is continuous. The interval is compact, so its quotient $X_A$ is compact. The unit circle is Hausdorff. The compact-to-Hausdorff continuous bijection theorem now shows that $\overline{F}$ is a homeomorphism:Indeed the theorem makes the inverse continuous because images of closed subsets of a compact space are compact and hence closed in a Hausdorff target.