Update: I asked my professor about this question and he said he forgot to add the property that $(X,d)$ is a complete meteric space with no isolated points. I was able to figure out the problem now.
I was working on a real analysis assignment with my friends and we were asked the following question:
Let $X$ be a non-empty set. Assume that $(X,d)$ has no isolated points and let $M = \{ A \subseteq X:$ either $A$ or $(X \setminus A)$ is countable $\}$ be a $\sigma$-algebra. Prove that $M$ does not contain the Borel $\sigma$-algebra.
We were looking at the set of rationals, and we thought $\mathbb{Q}$ has no isolated points and any open subset of $\mathbb{Q}$ is countable and therefore in $M$. So we thought $M$ would contain the Borel $\sigma$-algebra. Is this counter example wrong in any way?