The question arose while comparing the notions of compactness, countable compactness, local compactness, and "Lindelofness" in Hausdorff spaces. It is straightforward to show that compactness implies any of the other properties. I found ready counterexamples (I will be glad to provide them if asked) for all but one of the other possible implications, namely the question of whether countable compactness implies local compactness. The paper "On Countably Compact Nonlocally Compact Spaces" by T. B. Rushing shows that, in general, countable compactness does not imply local compactness. The examples he provides, however, are not Hausdorff.

Examples abound: take for instance a $\Sigma$product of twopoint spaces. To be specific let $X$ be the set of points in $\lbrace0,1\rbrace^{\omega_1}$ that have only countably many coordinates that are $1$. This set is dense but not open in the product, hence not locally compact but it is countably compact as each countably infinite subset sits in a compact subset of $X$. 


Le $X$ be any Hausdorff, sequentially compact, not compact space (e.g. $\omega_1$ with the order topology). Then $X^\mathbb{N}$ is Hausdorff, sequentially (hence countably) compact, and not locally compact, because any set with nonempty interior is mapped surjectively onto $X$ by some projection, thus is not compact as $X$ itself is not. 


Example 3.10.17 of Engelking's General Topology (Heldermann Verlag Berlin 1989) is the following dense subspace of $I^{\mathbb R}$, where $I=[0,1]$: Let $X$ be the set of all $(x_t)_{t\in\mathbb R}\in I^{\mathbb R}$ such that $x_t$ is different from zero for at most countably many $t\in\mathbb R$. Being a subspace of $I^{\mathbb R}$, the space is Hausdorff. The space is not locally compact: Hence $X$ is not locally compact. 


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), 586590) 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. 

