Preregular and locally compact implies regular The title says it all, and it was motivated from the following entry in Wikipedia:

''There are many situations where another condition of topological spaces (such as normality, pseudonormality, paracompactness, or local compactness) will imply regularity if some weaker separation axiom, such as preregularity, is satisfied.'' 

I don't know if preregularity is standard (since I couldn't find it in the classical books), so I will define it here:

Definition: A topological space $X$ is called preregular if whenever two points $x$ and $y$ do not share all their neighbourhoods, they have two disjoint neighbourhoods.


Some more motivation: There are two definitions for local compactness: (Kelley) every point admits at least one compact neighbourhood; or (Willard) if every point admits a local basis consisting of compact neighbourhoods.
Wikipedia uses the definition of local compactness in the sense of Kelley. If a space is locally compact in this sense and regular, then it is locally compact in the sense of Willard. A positive solution for the question allows us to weaken regularity by preregularity.

I worked on top of this, but I didn't  get anywhere worth explaining.
 A: The notion of Kolmogorov quotient is useful here. On a space $X$ we can define an equivalence relation $x\sim y$ if $x$ and $y$ share all neighborhoods, or equivalently, if $\overline{\{x\}}=\overline{\{y\}}$. Basically, this identifies all points which are topologically indistinguishable, and the quotient $K(X)=X/\sim$ is a $T_0$ space. The quotient map $q:X\to K(X)$ has every nice property one could hope for, in particular, the induced functions $q:\mathcal P(X)\to\mathcal P(K(X))$ and $q^{-1}:\mathcal P(K(X))\to\mathcal P(X)$ are mutually inverses on the topologies of $X$ and $K(X)$.
Now preregular spaces can be characterized as those spaces whose Kolmogorov quotient is Hausdorff. For regularity, one even has an equivalence: $X$ is regular if and only if $K(X)$ is regular. This is also true for other properties: Every point in $X$ has a compact neighborhood iff every point in $K(X)$ does. And $X$ is locally compact in the strong sense iff $K(X)$ is.
We can use this to proof your claim. Let $X$ be a preregular space such that every point has a compact neighborhood. Then its Kolmogorov quotient $K(X)$ is a Hausdorff space with the same property, and thus regular. It follows that $X$ is regular as well.
