This recent question led to some discussion on hemicompactness. A topological space $X$ is said to be hemicompact if there is an increasing sequence of compacta $K_1\subseteq\dots \subseteq K_n\subseteq \dots$ such that every compact $K\subseteq X$ lies in some $K_n$.
This is always the case when $X$ is $\sigma$-compact and weakly locally compact (every point has a compact neighborhood), and the converse is true under the additional assumption of first countability (since a hemicompact space is trivially $\sigma$-compact, the nontrivial part of the converse is the deduction of weak local compactness).
On the other hand, here it was shown that we cannot relax the first countability assumption to sequentialness. The counterexample is the Arens space, which is sequential (though not Fréchet-Urysohn) and hemicompact, yet not weakly locally compact. It should be noted that the Arens space has rather good separation properties ($T_6$).
A natural question becomes, given that sequentialness is too weak, can we relax first countability to the Fréchet-Urysohn property?
The short answer is no, at least not without further assumptions. I'll describe a counterexample next, then conclude with the titular question:
Counterexample.
Let $X=\mathbb R\cup \infty$, where $\mathbb R$ has the Euclidean topology and neighborhoods of $\infty$ are those of the form $\{\infty\}\cup\mathbb R\backslash D$, where $D\subset \mathbb R$ is discrete and closed (equivalently, $D$ has no limit points in $\mathbb R$).
Lemma. $K\subseteq X$ is compact if and only if either $K\subseteq \mathbb R$ is Euclidean compact, or $\infty\in K$ and $K\cap \mathbb R$ is bounded.
Proof.
If $K\subseteq \mathbb R$, then since $\mathbb R$ has the Euclidean topology $K$ is compact if and only if $K$ is Euclidean compact. On the other hand, if $\infty\in K$, and $K\cap \mathbb R\subseteq [-M,M]$, then every open cover of $K$ has a member $U\ni \infty$, whereby $\mathbb R\backslash U$ is closed and discrete, so that $K\backslash U\subseteq [-M,M]\backslash U$ is finite.
Finally, if $K\cap \mathbb R$ is unbounded, then let $x_n\in K\cap \mathbb R$ be a sequence eventually leaving every bounded subset of $\mathbb R$. Then the sets $D_n=\{x_k\mid k\geq n\}$ are discrete and closed, and so the sets $U_n= \{\infty\}\cup\mathbb R\backslash D_n$ form an open cover of $K$ with no finite subcover.
Corollary. $X$ is hemicompact, Fréchet-Urysohn, and not weakly locally compact.
Proof. Every neighborhood of $\infty$ intersects $\mathbb R$ in an unbounded set, so there are no compact neighborhoods of $\infty$, hence $X$ is not weakly locally compact. On the other hand, the sets $K_n:=[-n,n]\cup\{\infty\}$ are an increasing sequence of compacta in $X$, and every compact $K\subset X$ lies in some $K_n$, so $X$ is hemicompact.
Finally, to verify the Fréchet-Urysohn property, since every point in $\mathbb R$ has a countable basis, it suffices to consider $S\subseteq \mathbb R$ with $\infty\in \overline{S}$. In this case, $S$ must have some limit point $x\in \mathbb R$ (otherwise $\{\infty\}\cup \mathbb R\backslash S$ is a neighborhood of $\infty$). But then if $s_n\in S$ is a (distinct) sequence with $s_n\to x\in\mathbb R$, then $s_n$ eventually leaves every closed discrete subset of $\mathbb R$, whereby $s_n\to\infty$ as well.
Question.
Now, we could let the matter rest there, but compared to the Arens space, the space $X$ constructed above has rather poor separation properties: it is $T_1$, but just barely, in the sense that it is not even $US$ (convergent sequences do not have unique limits, as seen from the proof above).
Is there a Hausdorff counterexample, i.e., is there a Fréchet-Urysohn hemicompact Hausdorff space that fails to be locally compact*? More generally, if a space is hemicompact and Fréchet-Urysohn, might any separation axioms, anywhere from $US$ through to $T_6$**, guarantee weak local compactness?
(* Note that in the Hausdorff case, weak local compactness and local compactness coincide – the latter term indicating the existence of a basis of compact neighborhoods at each point.)
(** See here for an in depth description of a chain of properties between $T_2$ and $T_1$, with the weakest being the aforementioned $US$.)
Best Answer
The sequential fan with $\omega$ many spines is a counter-example:
It is defined as follows:
Let $X = \omega \times (\omega+1)$ with the product topology, where both factors have the order topology, i.e., the first factor is discrete, the second one is a convergent sequence.
Identify all points $(n, \omega)$ to a point called $\infty$.
The sequential fan with $\omega$ many spines is the quotient space $S = S(\omega) = (\omega \times \omega) \cup \{\infty\}$.
Each singleton except $\{\infty\}$ is closed and open in $S$, hence $S$ is T2, regular and thus even perfectly normal (since it is countable).
Neighborhoods of $\infty$ are of the form $\{\infty\} \cup \bigcup_{n \in \omega} (\{n\} \times [m_n, \omega))$.
Obviously, each $K_n := ([0, n] \times \omega) \cup \{\infty\}$ is compact (since it is the image of a compact set in $X$).
Moreover, by the above it is easy to see: