Why is the real line not used in Descriptive Set Theory?

There are several useful properties of $\mathbb{N}^\mathbb{N}$:

• $\mathbb{N}^\mathbb{N}$ is homeomorphic to $\mathbb{N}^\mathbb{N}\times\mathbb{N}^\mathbb{N}$. So we can view each element of $\mathbb{N}^\mathbb{N}$ as a code for a pair of elements, and the decoding maps are continuous. This is not the case for $\mathbb{R}$.
• $\mathbb{N}^\mathbb{N}$ is not connected. Because $\mathbb{R}$ is connected, there are no nonconstant continuous functions from $\mathbb{R}$ to $\{0,1\}$; you have to move up a few levels in the Borel hierarchy to get such functions. This doesn't matter if you just care about the functions being Borel, but when you're looking at the actual levels of the Borel hierarchy it's more convenient to start with continuous functions rather than starting a little higher up.
• It's easy to construct an element of $\mathbb{N}^\mathbb{N}$: you just have to construct a sequence of natural numbers. On the other hand, the representations of real numbers (Cauchy sequences, Dedekind cuts) are not as straightforward to work with. That makes proofs technically more difficult without making them more interesting. For example, compare the diagonalization proof that $\mathbb{N}^\mathbb{N}$ is uncountable with the diagonalization proof that $\mathbb{R}$ is uncountable.

There are also a few reasons that the use of $\mathbb{N}^\mathbb{N}$ does not result in a loss of generality:

• For any uncountable complete separable metric space (c.s.m.s.) $X$, there is a bijection between $X$ and $\mathbb{N}^\mathbb{N}$ that is both Borel measurable and has a Borel measurable inverse. So if the property we are studying is preserved by Borel isomorphisms, we can just replace an uncountable c.s.m.s. $X$ with $\mathbb{N}^\mathbb{N}$.

• Every c.s.m.s. is a continuous image of $\mathbb{N}^\mathbb{N}$. In fact, for any c.s.m.s. $X$ there is a closed subset $C$ of $\mathbb{N}^\mathbb{N}$ and a continuous bijection from $C$ to $X$. So if we are studying a property preserved by continuous maps, we can work with $\mathbb{N}^\mathbb{N}$ or with its closed subspaces without losing generality.

Those types of reasons are why it's safe to stick with $\mathbb{N}^\mathbb{N}$ most of the time: the goal is to study an arbitary c.s.m.s. (including $\mathbb{R}$) but for most purposes there's no loss of generality in studying $\mathbb{N}^\mathbb{N}$.