Let X be a topological space that is not too bad (let's say "not too bad" = "compactly generated Hausdorff"), and let ∼ be an equivalence relation such that X /∼ is compact Hausdorff.
Does there exist a compact subspace A⊂X that meets every equivalence class of ∼?
(This would then imply that A /∼ is homeomorphic to X /∼).
The answer is positive if $X$ is second countable and locally compact, and $X/\sim$ is first countable (in addition to being compact Hausdorff). Proof: we claim that any point in $X/\sim$ has a neighborhood which is contained in the image of a compact subset of $X$. Given this, the rest is easy: use compactness of $X/\sim$ to find a covering of it by finitely many such neighborhoods and take the union of the compact sets whose images contain them. (This is Agol's technique.)
To prove the claim, suppose it fails and let $x \in X/\sim$ be a falsifying point. Fix a countable base $(U_n)$ of $X$ (second countability) and wlog assume each $U_n$ is precompact (local compactness). By first countability of $X/\sim$, we can now find a sequence $(x_n)$ in $X/\sim$ that converges to $x$ and such that $x_n$ is not in the image of $U_1 \cup \cdots \cup U_n$. Since $x$ is contained in the image of some $U_n$, eventually $x_n \neq x$, so wlog we can assume $x_n \neq x$ for all $n$. Now let $C$ be the set of points in $X/\sim$ whose image is one of the $x_n$. This set cannot be closed, for then its complement would be open and $(x_n)$ could not converge to $x$. Therefore it must not contain some boundary point $\bar{x}$, and this point must map onto $x$. Finally, by local compactness of $X$ some $U_n$ must contain $\bar{x}$, which contradicts the choice of the sequence $(x_n)$. We conclude that the claim must hold.
Here is a counterexample, inspired by Henrik Rüping's answer, but somewhat simpler:
Take $X$ to be the disjoint union of
$\quad X_1:=([0,1]\times [0,1])\setminus \{(0,0)\}$
and
$\quad X_2:=(]0,1]\times [0,1])\cup ([-1,0]\times\{0\})$.
Both $X_1$ and $X_2$ are given the subspace topology from $\mathbb R^2$.
The quotient space is $[0,1]^2$, with quotient map $\pi:X\to [0,1]^2$.
The map $\pi|_{X_1}$ is the inclusion $X_1\hookrightarrow [0,1]^2$.
The map $\pi|_{X_2}$ is a continuous bijection $X_2 \to [0,1]^2$ given by $\pi(-a,0)= (0,a)$ for $(-a,0)\in [-1,0]\times\{0\}$.
Here's a non-separable counterexample:
Let $D^2:=\{(\theta,r)|\theta\in S^1, r\in[0,1]\}/\sim$ be the unit disc in $\mathbb R^2$, parametrized in polar coordinates.
Let's call a subset $K\subset D^2$ thin if $0\in K$, it is compact, and for every convergent sequence $(\theta_n,r_n)\to 0$ of elements of $K$, the limit $\lim\theta_n\in S^1$ exists.
Let $X$ be disjoint union of all thin subspaces of $D^2$. Then $X$ maps surjectively onto $D^2$, and this is a quotient map.
However, no compact subspace of $X$ maps surjectively onto $D^2$. A compact subspace of that infinite disjoint union is necessarily contained in a finite disjoint union, and the union of finitely many thin subspaces cannot be the whole of $D^2$.
If $X$ is locally compact, and the map $X \to X/\sim$ is an open mapping, then I think this holds. Consider a collection of open subsets $\cup_i U_i=X$, such that $\overline{U_i}$ is compact. Then the image of $U_i$ in $X/\sim$ is open by hypothesis. Choose finitely many $U_i$'s whose images cover $X/\sim$, then the union of their closures gives a compact subset mapping onto $X/\sim$.
