= Solution
\b[No countable subset works.] For any <countable set> $A\subseteq\mathbb R$, choose $b_n\in(n,n+1/2)\setminus A$: each nonempty interval is <uncountable>. The sequence tends to infinity, so every tail
$$
K_n=\{b_k:k\geq n\}
$$
is <closed> in $\mathbb R$ and disjoint from $A$. By part (i), $p(\mathbb R\setminus K_n)$ is open in $X_A$. These sets increase and cover $X_A$, since no point belongs to every tail. But any finite selection is contained in its largest member, which still misses every quotient point $p(b_k)$ in that member's remaining tail. The $b_k$ represent distinct singleton classes.
This is an <open cover> without a <finite subcover>, proving
$$
\boxed{A\text{ countable}\ \Longrightarrow\ \mathbb R_A\text{ is not compact}.}
$$
The argument does not assume that $A$ is closed, bounded, or discrete. It is the <noncompactness after collapsing a countable real subset> obstruction.
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