 Noncompactness after collapsing a countable real subset

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.