A space $X$ is called core-compact if the set of all open set in $X, \mathcal{O}(X)$, is a continuous poset.
It is known that every locally compact is core-compact.
Here, a space $X$ is locally compact if every $x\in X$ and every neighbourhood of $x$ contains a compact neighbourhood of $x$.
I am looking an example of core-compact but not locally compact.
I am thinking $\mathcal{R}$ with co-countable topology. This space is not locally compact.
For any $U\in\mathcal{O}(R)$, $U\neq\sup\{V: V\ll U\}$, where $V\ll U$ means every open cover $\{U_i\}$ of $U$ there is a finite subcover of $V$. Indeed, since for every $U\in\mathcal{O}(R)$, $\{V: V\ll U\}=\emptyset$.
I only saw this question today; it is indeed hard to find examples of this sort of thing.
A somewhat elaborate example, essentially the spectrum of the distributive continuous lattice of lower semicontinuous functions $I \to I$ on the unit interval (where elements of the spectrum are prime elements of the lattice), is given in The Spectral Theory of Distributive Continuous Lattices by Hofmann and Lawson, section 7.
Unfortunately, your co-countable topology example isn't core-compact. Indeed, for the maximal open $\mathbb{R}$ in this topology, there is no nonempty open $W$ satisfying $W \ll \mathbb{R}$. For, such $W$ contains an infinite countable set $A = \{a_1, a_2, a_3, \ldots \}$, and then the sets $\{a_n\} \cup (\mathbb{R} \setminus A)$ form an open cover of $\mathbb{R}$ with no finite subcover of $W$. This example appears on page 9 of this report.