compact and locally Hausdorff, but not locally compact

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 $q$.)