Not $\sigma$-compact set without axiom of choice 
Today in measure theory, we introduced the concept of a $\sigma$-compact set, which is a set which can be expressed as the countable union of compact sets.

Since the set of $\sigma$-compact sets seemed to be very large (I couldn't think of a not $\sigma$-compact set), I asked for a set which is not $\sigma$-compact, but our instructor couldn't come up with a set, which he didn't have to explicitly construct with the axiom of choice, for which we didn't have time.

My question is if there is such an set which can be constructed without the axiom of choice.

 A: The axiom of choice is not needed to construct a subset of $\Bbb R$ which is not $\sigma$-compact (which I guess is ultimately what this question is about).
The irrational numbers form such a set, since any compact subset of $\Bbb{R\setminus Q}$ has an empty interior, and therefore any $\sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.
(To the initiated, Baire's theorem is provable in $\sf ZF$ when the space is separable, like $\Bbb{R\setminus Q}$ which has $\pi+\Bbb Q$ as a dense subset.)
A: Note that - in $\mathbb{R}$ at least - a set is $\sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_\sigma$ (or $\Pi^0_2$). A fundamental theorem concerning the topology of $\mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_\sigma$. 
So in particular, the set of irrationals is not $\sigma$-compact.
