Consider $\mathbb{R}_\ell$ be the the 'Sorgenfrey line':
Real line with the topology constructed from the intervals $\{[a,b):a<b\}$.
Prove that $\mathbb{R}_\ell$ is not locally compact.
I show here that every compact subset of $\mathbb{R}_\ell$ is countable, so has empty interior in particular. So there are no non-empty open sets inside of some compact set, so whatever your definition of local compactness, $\mathbb{R}_\ell$ fails it.
Suppose it is. Then every $x$ has a compact nbd $K$ such that $x \in [a,b) \subseteq K \subseteq \mathbb R$. Since $\mathbb R_l$ is $T_2$ so $K$ is a compact, $T_2$ space. Since $[a,b)$ is closed in $\mathbb R_l$, so $[a,b) \cap K=[a,b)$ is closed in $K$. But in a compact $T_2$ space, a set is closed iff compact. So $[a,b)$ is compact in $K$ and thus in $\mathbb R_l$, which is absurd.
Suppose it is. Then every $x$ has a compact nbd $K$ such that $x \in [a,b) \subseteq K \subseteq \mathbb R$. But for a suitable $n$, $$(-\infty,a) \cup\big(\bigcup_{i=n}^{\infty}[a,b-\frac{1}{i})\big)\cup[b,\infty)$$ is an open cover of $K$ with no finite subcover, which is absurd.