The definition of a limit point I will use here is the one provided in baby Rudin:
$l\in E$ is a limit point of $E$ if for every open nbhd $U$ of $l$, $(U\setminus \{l\}) \cap E\not= \varnothing.$
My argument for the statement in the title is the folloing:
Take an irrational $i\in \mathbb{R}\setminus \mathbb{Q}=\mathbb{Q}^c$. For it to be a limit point (of $\mathbb{Q}$), we need to show that for every nbhd $U$ of $i$, $(U\setminus \{i\})\cap \mathbb{Q}\not=\varnothing.$
And this can be done by choosing $r_n\in\mathbb{Q}$ such that $r_n\in (i-\frac{1}{10^{(n+1)}},i+\frac{1}{10^{(n+1)}})=U_n$ so that $r_n\in(U_n\setminus \{i\})\cap \mathbb{Q}$ for each $n\in\mathbb{N}$ (since every open nbhd $U$ of $i$ should contain $U_n$ for some $n$). i.e., $r_n$ has length $n$ decimal tail which is equal up to $n$-th decimal of $i$. For example, if $i=0.23554245...$, then $r_5=0.23554$.
But, according to this, all rationals can be limit points of $\mathbb{Q}$ too. With slightly different choice of $r_n$, e.g. for a rational $1$, $r_n=0.9999...9$ up to $n$-th decimal. So every point of $\mathbb{Q}$ and $\mathbb{Q}^c$, namely $\mathbb{R}$, is a limit point of $\mathbb{Q}$.
Is my argument valid? If it is, is it rigorous enough?