Even worse! ;) You might have a Hausdorff countably compact space
which is not locally countably compact in the stronger sense that every
point has a neighborhood base of countably compact sets (of course, for Hausdorff spaces, the strong and weak forms of "locally compact" are equivalent; but you need $T_3$ to get the corresponding equivalence for "locally countably compact").
Consider $ omega_2$, the third infinite cardinal, with the usual order topology, and make the topology finer by adding $A= { alpha in omega_2 | alpha text{ has cofinality } omega_1 }$ as an open set. So, a base for the topology is given by the open intervals together with the sets of the form $L cap A$, for $L$ an open interval.
With this topology, $ omega_2$ remains countably compact, since every infinite
subset has a complete accumulation point. It obviously remains Hausdorff. Notice that, with this topology,
any ordinal of cofinality $ omega_1$ is an isolated point,
provided it is not a limit of a set of ordinals of cofinality $ omega_1$.
If you consider instead an ordinal $alpha$ of cofinality $ omega_1$ such that $alpha$ is also a limit of ordinals of cofinality $ omega_1$, then $A$ is a neighborhood of $alpha$, but no neighborhood $U$ of
$alpha$ contained in $A$ is countably compact, since any such $U$ contains
a closed infinite discrete set, by the above remark (I mean, the infinite discrete set is closed in U).
I am sure that this example or something similar appears in the literature, but I do
not recall where. Probably there are also simpler counterexamples (in the counterexample above, too, you can start with an ordinal much smaller than $ omega_2$, of course). I just found that N. Noble (Two examples on preimages of metric spaces , Proc. Amer. Math. Soc. 36 (1972), 586-590) mentions that the existence of
such counterexamples is relevant in constructing other counterexamples about $k$-spaces.
A lot of time has gone by, but I think it might be useful to point this out.
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