Let $X = \mathbb{R}$ with standard topology, $A \subseteq R$ and define an equivalence relation $\mathcal{R}_A$ on $X$ with classes $[x]=\{x\}$ if $x \notin A$ and $[x]=A$ if $x \in A$. Is there a countable subset $A \subseteq \mathbb{R}$ such that the quotient topological space $X/\mathcal{R}_A$ is compact?
Here is my attempt:
No. Suppose $A \subseteq \mathbb{R}$ is countable, then we enumerate the elements as $A=\{a_1,a_2,...\}$ and for $\epsilon > 0$, let
$$\mathcal{A}:=\{(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i}) \mid i \geq1\}$$
be an open cover for $A$. If we let $p:X \rightarrow X_A$ be the quotient map then
$$X_A \setminus \{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\}$$
is non-empty because $\mathcal{A}$ does not cover $\mathbb{R}$ by considering the sum of measures
$$\Big| \bigcup (a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i}) \Big| \leq \sum \Big|(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big|< \infty.$$
But we have homeomorphism
$$X_A \setminus \{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\} \cong \mathbb{R}\setminus \bigcup(a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})$$
and since the set on the RHS is not bounded, it is not compact, so we can find an open cover for RHS (hence LHS) without a finite subcover. Then the open cover for LHS together with $\{p\Big((a_i-\frac{\epsilon}{2^i},a_i+\frac{\epsilon}{2^i})\Big) \mid i \geq1\}$ is an open cover for $X_A$ which has no finite subcover. $\square$