I say that a space $S$ is weakly locally compact if for every $x\in S$ there is an open set with $x\in V$ with $\overline{V}$ compact.
Our teacher gave us this exercise, which is similar to previous math stack exchange exercises, except for the fact we do not ask of $\mathcal{M}$ to be Hausdorff:
If $\mathcal{M}$ is weakly locally compact prove that if $A\subset \mathcal{M}$ is closed it must be weakly locally compact. Also prove that if $A\subset \mathcal{M}$ is open it must be weakly locally compact.
I was only able to prove this when $A$ is closed:
Let $x\in A$. There is an open set such that $x\in V$ and $\overline{V}$ is compact. $V\cap A$ is an open neighborhood for $x$ (in the subspace topology). Furthermore, we have that the closure of this set (in the subspace topology) is:
$$A\cap \overline{V\cap A}\subset A\cap \overline{V}\cap \overline{A}=A\cap \overline{V}\subset \overline{V}$$
Clearly, closed subsets of compact sets are compact (we can create an open cover of the larger compact set using the complement of the smaller set). Using this, we have that $A\cap \overline{V} $ is compact and consequently, $A\cap \overline{V\cap A}$ is also compact!
This finishes the proof. It is very hard to adapt this to $A$ being open, because I cannot find a finite subcover...
Is there a nice counterxample when A is open?
The Countable complement topology on the real numbers fails to be weakly locally compact. (Though the pi-Base uses the definition that each point has a compact neighborhood: for each $x$ there's open $U$ and compact $K$ with $x\in U\subseteq K$. Note that without $T_2$, compacts may not be closed, so $\overline{U}$ may not be a closed subset of $K$ and thus may not be compact. However, this particular space does have the property that compacts are closed even though it's not $T_2$, so we avoid this possible technicality.)
So let $X=\mathbb R\cup\{\infty\}$ where $X$ is the only neighborhood of $\infty$ and the subspace $\mathbb R$ has the countable complement topology. This is a $T_0$ but not $T_1$ space, it is weakly locally compact as $X=\overline{X}$ is a compact neighborhood of each point, but its open subspace $\mathbb R$ is not weakly locally compact.