This question follows on from a previous one, which has been answered in the negative: Is a locally compact Hausdorff quotient of a locally compact $\sigma$-compact first countable Hausdorff space always first countable?
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 $X$ a Frechet-Urysohn space?
Best Answer
$\newcommand{\int}{\operatorname{int}}$ I realize this is a little old but since it hasn't been resolved, and I was thinking about similar issues recently:
Yes, $X$ must be Fréchet-Urysohn.
Proof. Suppose otherwise, and choose $x_n\in K\backslash q(\int(L_n))$ for each $n$. Let $V_n=q^{-1}\left(X\backslash \bigcup_{k\geq n}\{x_k\}\right)$. Note that for each $m$, only finitely many of the closed sets $q^{-1}(\{x_k\})$ intersect $\int(L_m)$, so $V_n\cap \int(L_m)$ is open. Since this holds for all $m$ and $\{L_m\}$ is an exhaustion, $V_n$ is open. But the sets $V_n$ are also saturated, so $\{q(V_n)\}$ is an open cover of $K$ with no finite subcover, contradicting compactness.
Proof of main result.
Suppose $x\in \overline{A}\subseteq X$. Let $K\ni x$ be a compact neighborhood of $x$. By the lemma, we have $K\subseteq q(\int(L_n))$ for some $n$. We claim that for any open $V\supseteq q^{-1}(\{x\})$, there is some open set $U\ni x$ with $$q^{-1}(\overline{U})\cap L_n\subseteq V.\tag{*}$$
To see this, suppose otherwise, and consider the family of closed sets $\{q^{-1}(\overline{U})\cap L_n\backslash V\}$, where $U$ is neighborhood of $x$. If these are nonempty, then they have the finite intersection property, hence by compactness of $L_n$, their intersection is non-empty. But by the Hausdorff condition on $X$, the intersection of this family is empty, as we can always find neighborhood $U$ of $x$ with $x'\notin \overline{U}$, and then $q^{-1}(\{x'\})\cap q^{-1}(\overline{U})=\emptyset$, so the intersection can only contain points in $q^{-1}(\{x\})$, and these are also excluded as they lie in $V$.
Therefore, given any open $V$ as in the preceding paragraphs, there is such a $U$. Since $x\in \overline{A}$ and $K\cap U$ is a neighborhood of $x$, $A\cap K\cap U\neq \emptyset$. Because $$q(\int(L_n))\supseteq K\supseteq A\cap K\cap U\neq \emptyset,$$ there is some $$y\in q^{-1}(U)\cap \int(L_n)\cap q^{-1}(A)\subseteq V\cap q^{-1}(A),$$ with the inclusion following from (*). Thus every neighborhood $V$ of $q^{-1}(\{x\})$ intersects $q^{-1}(A)$, and so some $y\in q^{-1}(\{x\})$ lies in $\overline{q^{-1}(A)}$. By the Fréchet-Urysohn property of $Y$, we have a sequence $b_n\in q^{-1}(A)$ with $b_n\to y$, and therefore $q(b_n)\in A$ is a sequence with $q(b_n)\to q(y)=x$.
Remarks.
As can be seen from examining the proof, the lemma does not require any separation on $Y$, and only $T_1$ for the quotient $X$. $Y$ need only be exhaustible by compacts (equivalently, $\sigma$-compact and weakly locally compact, so that every point has a compact neighborhood).
The proof of the main result itself does require $X$ to be locally compact Hausdorff, though $Y$ need only be exhaustible by compacts and Frechet-Urysohn.