$\mathbb{R}$ and $\mathbb{Q}$ are $\sigma$-compact, and Lindel"of assuming countable choice; $\mathbb{R}$ is locally compact and $\mathbb{Q}$ is nowhere locally compact

Each $I_n$ is closed in $\mathbb{R}$, its complement being the union of the open sets $\{t : t < -\iota(n)\}$ and $\{t : t > \iota(n)\}$, and it is bounded; so $I_n$ is a compact subset of $\mathbb{R}$ by [L2]. By [L3] every real $t$ satisfies $|t| < \iota(n)$ for some $n$, so $\mathbb{R} = \bigcup_{n \in \mathbb{N}} I_n$, an at most countable union of compact subsets: claim 1.

Each singleton $\{r\}$ with $r \in \mathbb{Q}$ is a compact subset of $\mathbb{Q}$, the subspace it carries being a one-point space; and $\mathbb{Q}$ is the union of the family of its singletons, which is at most countable by [L4]. So $\mathbb{Q}$ is $\sigma$-compact: claim 2.

For claim 4 in $\mathbb{R}$: given $p \in \mathbb{R}$ the set $\{t : |t-p| \le 1\}$ is closed and bounded, hence compact by [L2], and it contains the open $(p-1,p+1) \ni p$, so it is a compact neighbourhood of $p$ and $\mathbb{R}$ is locally compact.

For claim 3 assume countable choice and let $\mathcal{U}$ be an open cover of $\mathbb{R}$. For $n \in \mathbb{N}$ the set $T_n$ of finite subfamilies of $\mathcal{U}$ covering $I_n$ is nonempty, $I_n$ being compact by step 1.1 and the ambient reading being licensed by [L9], so [L5] supplies $\mathcal{V}_n \in T_n$ for every $n$; the union $\bigcup_{n \in \mathbb{N}} \mathcal{V}_n$ is an at most countable subfamily of $\mathcal{U}$ by [L10], being a countable union of finite sets, and covers $\mathbb{R}$ by step 1.1. The same argument with the singletons of step 1.2 in place of the $I_n$ shows $\mathbb{Q}$ is Lindelöf: claim 3.

For claim 4 in $\mathbb{Q}$, let $r \in \mathbb{Q}$ and suppose $K \subseteq \mathbb{Q}$ were a compact neighbourhood of $r$ in $\mathbb{Q}$; then some set open in $\mathbb{Q}$ lies between $r$ and $K$, so by [L1] there is a real $\varepsilon > 0$ with $(r-\varepsilon, r+\varepsilon) \cap \mathbb{Q} \subseteq K$, and by [L8] the set $K$ is a compact subset of $\mathbb{R}$ as well, hence closed in $\mathbb{R}$ by [L2].

By [L7] there is an irrational $t$ with $r < t < r + \varepsilon$. Every neighbourhood of $t$ contains an interval $(c,d)$ with $r < c < t < d < r+\varepsilon$, and [L7] puts a rational $q$ with $c < q < d$ in it; that $q$ lies in $(r-\varepsilon, r+\varepsilon) \cap \mathbb{Q} \subseteq K$. So every neighbourhood of $t$ meets $K$, and $K$ closed gives $t \in K \subseteq \mathbb{Q}$ by [L8], contradicting the irrationality of $t$. Hence no point of $\mathbb{Q}$ has a compact neighbourhood in $\mathbb{Q}$, which completes claim 4.