I was given a "proof" that contradicts the Baire Category Theorem and I can't figure out why it doesn't work.
Let $C$ be the Cantor set in $[0,1]$. Then $[0,1]\setminus C$ is a disjoint union of countable open intervals. Take a point from each of these sets and call the set of such points $A$. Define $X = C \cup A$, which is closed, so it is a complete metric space and thus the Baire Category Theorem holds in $X$. Then $A$ is open in $X$, since it is the intersection of $X$ and the complement of $C$. Additionally, $A$ is countable, so it is first category. Then $A$ is an open set of first category, which contradicts the Baire Category Theorem.
I suspect that $X$ is not actually closed ($A$ certainly isn't closed), but I am having trouble proving that is the case i.e. finding a limit point of $X$ not contained in $X$. I think the way to do that would be to construct a sequence using both $C$ and $A$ whose limit isn't in $A$ or $C$, but I am not sure how to go about doing that.
Each point of $A$ is an isolated point of $X$, so $A$ is not the union of countably many nowhere dense sets in $X$: if $x\in A$, $\operatorname{cl}_X\{x\}$ has non-empty interior, namely, itself. Thus, $A$ is not first category in $X$.