Let $\mathbb{R}$ be a topological space with topology consisting of the sets $A \cup B$, where $A$ is open in the usual topology, and $B \subseteq \mathbb{R} \setminus \mathbb{Q}$. Is the interval $[0,1]$ compact in this topology?
I think that it might be, as if we have an open cover of $[0,1]$ in this topology, the cover consists of open sets of the form $A \cup B$. $B$ is a set of irrational numbers in the interval $[0,1]$. We can always find an open interval in $[0,1]$ to cover each irrational number, since for any irrational $p \in [0,1]$ there are $a,b \in \mathbb{Q} \cap [0,1]$ such that $a < p < b$, and then $p \in (a,b)$. Therefore now we have an open cover entirely composed of sets from $A$. But we know closed intervals in the usual topology are compact, so then $[0,1]$ is compact in the $A \cup B$ topology.
Could you please tell me whether my argument is correct or not?