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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 since the inverse image of each $x_n$ is closed this point $\bar{x}$ 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.

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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 $U_1 \cup \cdots \cup U_n$. By local compactness of $X$, any point in Since $X/\sim$ x$is contained in the image of some$U_n$; applying this comment to$x$shows that U_n$, 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 since the inverse image of each $x_n$ is closed this point $\bar{x}$ 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.

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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 $U_1 \cup \cdots \cup U_n$. By local compactness of $X$, any point in $X/\sim$ is contained in the image of some $U_n$; applying this comment to $x$ shows that 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 since the inverse image of each $x_n$ is closed this point $\bar{x}$ 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.