I have the following problem:
Suppose $A$ is a closed subset of a complete metric space $X$. Show that $A$ is of the first category if and only if $A$ has empty interior.
My attemp: if $A$ is a closed subset of $X$ and has empty interior, i.e., $int(A)=\emptyset$, we can conlude that $A=\overline{A}\implies int(A)=int(\overline{A})=\emptyset$. So, $A$ is nowhere dense.
Now I define the countable sequence $U_{n}$ equal to $A$ if $n=1$ and $\emptyset$ if $n>1$. Therefore, $A=\displaystyle\cup_{n=1}^{\infty}{ U_{n}}$ is the first category.
Now, if $A$ is a closed subset of $X$ and is of the first category, my idea was prove this by contradiction, suppose that a closed ball $Y$ is contained in $A$. Since $Y\subset X$, and $X$ is a complete metric space, we can consider $Y$ as a complete metric space, by Baire's theorem, $Y$ is the second category, and I don't know how continuous this. I need some idea, thanks!
If $A$ has non-empty interior, then $B[x; r]$ is contained in $A$, for some $x \in X$ and $r > 0$. So, if we can express $A$ as a countable union of nowhere dense sets $\bigcup_{n \in \mathbb{N}} C_n$, then $B[x; r]$ is covered by this countable union of nowhere dense sets. Consider $D_n = C_n \cap B[x; r]$, and you have that $B[x; r]$ is itself the countable union of nowhere dense sets $D_n$.
Of course, $B[x; r] \setminus \overline{D_n}$ is an open dense subset of $B[x; r]$, whose intersection is empty. This contradicts the BCT on $B[x; r]$.