X Hausdorff: every point has precompact neighborhood iff X has a basis of precompact open sets Claim Suppose $X$ is Hausdorff, then:
Every point $p\in X$ has a precompact neighborhood in $X$ $\Longleftrightarrow$ $X$ has a basis of precompact open subsets
Question: It's not clear to me why it's necessary for X to be Hausdorff; so, what is missing from the following proof that X being Hausdorff provides?
Proof:
"$\Longrightarrow$"
Suppose that every point $p \in X$ has a precompact neighborhood in $X$. We will construct a basis using precompact open subsets of $X$.
Denote the set of precompact neighborhoods for each point as $\mathbb{U} = \left\{ U_p \right\}_{p\in X}$. Denote the topology of $X$ as $\tau$. We claim that $\forall u \in \tau$ and $\forall U_p \in \mathbb{U}$, the intersection, $u_p = u \cap U_p$ is precompact. This follows from the fact that every closed subset of a compact subspace is compact. Thus, $\overline{u}_p \subset \overline{U}_p$ implies that $\overline{u}_p$ is compact.
Therefore, the following is a precompact basis: $\mathcal{B} = \left\{u_p: \exists u \in \tau, \exists U_p \in \mathbb{U}\,\, \mathrm{ s.t. }\,\, u_p = u \cap U_p \right\}$
"$\Longleftarrow$"
Suppose $X$ has a basis of precompact open subsets, which we'll denote $\mathcal{B}$. $X$ is open, so $X =\underset{u\in \mathcal{B}}\cup u$. Thus, $p \in X$ implies that $\exists u \in \mathcal{B}$ such that $p \in u$. So, every $p$ has a precompact neighborhood.

$X=(\mathbb{R}\times\mathbb{Z})/\sim$, where $\sim$ is the equivalence relation generated by
$(x,n)\sim (x,m)$ for all $n,m\in \mathbb{Z}$ and all $x\ne 0$.
$X$ is non-Hausdorff. $X$ is locally compact but does not have a basis of precompact open subsets.  This comes from "Introduction to Topological Manifolds" by J.M. Lee problem 4-22.  The proof by Henno Bransdma above assumes the existence of a precompact neighborhood that does not exist for the origins of this set.
