An open cover $([0,1] \cap \Bbb{Q}) \times ([0,1] / \Bbb{Q})$ has no finite subcover?

Question

I was thinking about the correct statement

An open cover $([0,1] \cap \Bbb{Q}) \times ([0,1] \setminus \Bbb{Q}) = \{(x,y) \in [0,1] \times [0,1] : x \in \Bbb{Q} , y \notin \Bbb{Q}\} \subset (\Bbb{R}^2, d_{\infty})$ that doesnot have any finite subcover.

Here we see that $\Bbb{Q} \cap [0,1]$ has no finite subcover but this is is a product of the rationals in $[0,1]$ and the irrationals.

Answer 1

Though there are many covers of
$([0,1] \cap \Bbb{Q}) \times ([0,1] / \Bbb{Q})$
that have finite subcovers, there are lots that do not.

If every cover had a finite subcover, then that space would be compact. As compactness is preserved by continuous functions, such as the first projection, $[0,1] \cap \Bbb{Q}$ would be compact, which of course, it isn't.