Let $Y$ be a locally compact, $\sigma$-compact, first countable Hausdorff space and $q: Y\to X$ a quotient map with $X$ Hausdorff. Suppose that $X$ is locally compact. Is $X$ first countable?
I have spent a while hunting the literature for an answer but have not been able to find one. It works the other way: if $X$ is first countable then $X$ is locally compact, but what about this way?
Here is a counterexample. Let $Y=[0,1]\times[0,1]$ with the order topology of the lexicographic order (so $Y$ is first countable, compact, and Hausdorff). Let $X$ be the quotient of $Y$ that identifies $[0,1]\times\{0,1\}$ to a single point. This is a closed subset of $Y$, so the quotient is still compact Hausdorff. However, $X$ is not first-countable: it is the one-point compactification of a disjoint union of uncountably many copies of $(0,1)$ (the image of $[0,1]\times(0,1)\subset X$, which has the product topology where the first coordinate is discrete and the second has the usual topology), which is not first-countable at the point at infinity.