$T_1$ spaces and "rank" of a set For a $T_1$ space $X$ define $\text{rank}(A) = \min\{\kappa : A\text{ is union of }\kappa\text{ closed sets}\}$. Clearly $\text{rank}(A) \leq 1$ iff $A$ is closed and $\text{rank}(A) \leq \aleph_0$ iff $A$ is a $F_\sigma$ set.
It follows that $\text{rank}(\mathbb{R}\setminus\mathbb{Q}) \geq \aleph_1$ for $X = \mathbb{R}$. Can we find rank of irrationals in ZFC?
 A: The Handbook on Set-theoretic Topology, Chapter 3 (by Erik van Douwen), the integers and topology is entirely about all sorts of cardinals that can be defined that are $\ge \aleph_1$ and $\le \mathfrak{c}$ (the continuum). One such is the dominating number $\mathfrak{d}$, which is defined as the minimal cardinality of a cofinal subset of $\omega^\omega$ (the set of all functions from $\omega$ to $\omega$ in the coordinatewise order, $f \le g$ iff $\forall n \in \omega: f(n) \le g(n)$; a subset $B$ of a poset $(A,\le)$ is cofinal in $A$ iff $\forall a \in A:\exists b \in B: a \le b$).
Thm 5.1 says (among other things) that whenever $\kappa \le \lambda$ are regular uncountable cardinals, it is consistent (i.e. there is a model of ZFC such ) that $\mathfrak{c}=\lambda$ and $\mathfrak{d} = \kappa$.
We denote the irrationals by $\Bbb P$, as topologists often do, and
in Thm. 8.10c it is shown that $\operatorname{kc}(\Bbb P)$ (the compact covering number, the minimal number of compact subsets needed to cover $\Bbb P$) is equal to this $\mathfrak{d}$ (in fact, this is true for any non-$\sigma$-compact Polish space). As we can consider $\Bbb P$ to lie in a compact space (e.g. $[-\infty, +\infty]$), this also answers the question on how many closed sets we need to cover $\Bbb P$.
Combining these facts, we conclude that basically any regular uncountable cardinal could be the rank of $\Bbb R\setminus \Bbb Q$, e.g. $\aleph_{42}$ if you like, e.g. (of course the continuum will be at least that large).
