TITLE: Is a $\sigma$ - compact Hausdorff space normal?
QUESTION [7 upvotes]: Are $\sigma$ -compact Hausdorff spaces normal?

REPLY [4 votes]: There may be some confusion based on definitions.  Engelking's text, in particular, defines $\sigma$-compact spaces to be regular (T$_3$) spaces which are the countable union of compact subspaces.  With this Brian's (original) answer works.  (I cannot stress enough that his current answer works perfectly well.)
However if you remove regularity from the definition of $\sigma$-compactness, you can get counterexamples, as Brian mentions above.  Another example is the following:  Let $A = \{ \frac{1}{n} : n \in \mathbb{N} \}$, and give $\mathbb{R}$ the topology by declaring the open sets to be of the form $U \setminus B$ where $U \subseteq \mathbb{R}$ is open in the usual metric topology and $B \subseteq A$.

As this topology is finer than the usual metric topology, it follows that this space is Hausdorff.
It is $\sigma$-compact since you can cover $\mathbb{R}$ by countably many closed (bounded) intervals which each contain only finitely many elements of $A$.
However this space is not regular because $A$ is closed in the new topology, but there is no pair of disjoint open sets around $0$ and $A$, respectively.