We know that a Hausdorff, secound-countable, and locally compact space is sigma-compact. As I understand, however, the Hausdorff requirement is typically attached to avoid choosing a particular definition of local compactness (see for instance the comments on the following answer: https://math.stackexchange.com/a/57352/776333).
One relaxation of this requirement is provided by https://ncatlab.org/nlab/show/locally+compact+and+second-countable+spaces+are+sigma-compact, where the property in the title of this question is proved on the understanding that locally compact means "that every point has an open neighbourhood whose topological closure is compact". My question is whether this can be relaxed further to "every point has a compact neighborhood", which (in my understanding) is a weaker requirement.
It seems to me that this should be so and I provide an attempt at a proof — if the broader definition of local compactness is insufficient and the proof is wrong, I hope that this will allow someone to identify where exactly my confusion lies. Alternatively, it looks to me like my proof relies on the axiom of choice. Perhaps the narrower definition allows the proof to go through without this?
Conjectured theorem: Every locally compact (in the sense that every point has a compact neighbourhood) and second-countable space is $\sigma$-compact.
Attempted proof:
Let $X$ be a locally compact and second-countable space and let $N = \{ B_i \subset X \}_{i\in I}$ be a countable basis. We assume the axiom of choice throughout, and countable is taken to mean either countable or finite.
For every $x \in X$ there exists a compact neighborhood by local compactness. Choose one such neighbourhood $C_x$ for each $x$ and thus construct the set $\mathcal{C} = \{C_x\}_{x\in X}$. Then for each $C_x$ there exists some open set $U_x$ s.t. $x \in U_x \subset C_x$ and by the definition of a basis there exists some $B_j$ s.t. $x \in B_j \subset U_x \subset C_x$. Denote by $\beta_x$ the non-empty set of basis elements $\{x \in B_j \subset U_x\}_{j \in I}$ and define a set $\mathcal{B} = \{\beta_x\}_{x\in X}$.
Now define a choice function $f$ on $\mathcal{B}$ selecting an element from each $\beta_x$ and construct the image set $f[\mathcal{B}]$. The elements of $f[\mathcal{B}]$ cover $X$ by construction (since we have chosen at least one $B$ for each $x$) and $f[\mathcal{B}] \subset N$. For convenience, call the set of indeces present in $f[\mathcal{B}]$ by $J$ s.t. $J \in I$.
Then for each element $B_k \in f[\mathcal{B}]$ define a non-empty set $E_k = \{C : C \in \mathcal{C}, B_k \in C\}$ and the set $\mathcal{E} = \{E_k\}_{k \in J}$. Define a choice function $g$ on $\mathcal{E}$ and now consider $g[\mathcal{E}]$. This is an image of a countable set and thus countable. It covers $X$ by virtue of each member containing a member of the covering $f[\mathcal{B}]$ and each member is compact. This is the desired countable cover by compact sets. $\square$
Best Answer
Yes, it is true: it's enough that $X$ is Lindelöf: just take a countable subcover of cover by the compact sets with non-empty interior (or just take the open covers of interiors). Second countability is overkill.