Here’s a counterexample to the conjecture.
Let $Y=\Bbb N\times\Bbb N$, let $p$ and $q$ be distinct points not in $Y$, and let $X=Y\cup\{p,q\}$. Points in $Y$ are isolated. For each $k\in\Bbb N$ the set
$$B_p(k)=\{p\}\cup\left\{\langle m,n\rangle\in Y:n\ge k\right\}$$
is a basic open nbhd of $p$. For each $k\in\Bbb N$ the set
$$B_q(k)=\{q\}\cup\left\{\langle m,n\rangle\in Y:m\ge k\right\}$$
is a basic open nbhd of $q$. For convenience, for $n\in\Bbb N$ let $S_n=\{n\}\times\Bbb N$.
Each open nbhd of $p$ contains all but finitely many points of each $S_n$, and each open nbhd of $q$ contains all but finitely many of the sets $S_n$. Thus, every open set containing both $p$ and $q$ contains all but finitely many points of $Y$, and $X$ is compact. $Y$ is Hausdorff, as are the sets $B_p(k)$ and $B_q(k)$, so $X$ is locally Hausdorff. However, none of the sets $B_p(k)$ contains a compact nbhd of $p$, so $X$ is not locally compact. (Similarly, $X$ fails to be locally compact at $q$.)
Suppose $U$ is an open subset of a locally compact Hausdorff space $X$, $K\subset U$ and $K$ is compact. Then there is an open set $V$ with compact closure such that $$K\subset V\subset \bar{V}\subset U.$$
This is Theorem 2.7 in Rudin's Real and Complex analysis. To prove it first you need to show (the easy) fact that in a Hausdorff space compact subsets and points can be separated. More precisely, if $K\subset X$ is compact and $p\in X\setminus K$, then there exist disjoint open sets $V, W$ containing $K$ and $p$, respectively.
For your purposes you need a special case of this theorem where $K$ is a point in the open set $U$.
Best Answer
The discussion on local compactness on its Wikipedia page is quite relevant to this question. It discusses the three common non-equivalent (in general spaces) definitional variants of local compactness, and notes that these are all equivalent for Hausdorff ($T_2$) spaces, and this is part of the reason that Hausdorffness and local compactness are often assumed together (the property behaves better wrt operations and the combination implies Tychonoff (completely regular), and the Alexandroff compactification (one-point compactification) is better behaved, among others..)
The definition of the OP (in the link) is definition variant $2''$: every point has a local base of relatively compact neighbourhoods.
The Alexandroff compactification of $\Bbb Q$ is an example of a $T_1$ space that is locally compact in senses $1$ and $2$ (so also loc. compact according to the OP, as $2''$ is equivalent to $2$) but not Hausdorff. That example answers the title's question.