Show that $A$ cannot be embedded into any countable union of closed nowhere dense subsets in $X$ Let $X=A\cup B$ be a complete metric space. Assume that $B$ is a countable union of closed nowhere dense subsets of $X$.
Show that $A$ cannot be embedded in any countable union of closed nowhere
dense subsets in $X.$ 
What I have so far:
Suppose $f:A\to \bigcup_{n\ge 1}F_n$ is an embedding of $A$ into $\bigcup_{n\ge 1}F_n,$ where $F_n$ is closed and nowhere dense in $X.$
Then $$A=f^{-1}(\bigcup_{n\ge 1}F_n)=\bigcup_{n\ge 1}f^{-1}(F_n).$$
Since $F_n$ is closed in $X,$ it is closed in $\bigcup_{n\ge 1}F_n,$ and so $f^{-1}(F_n)$ is closed in $A.$ So we can write $f^{-1}(F_n)=C_n\cap A,$ where $C_n$ is a closed set in $X.$ It follows that $A\subset \bigcup_{n\ge 1}C_n,$ and so $X=\bigcup_{n\ge 1}C_n\cup B.$ 
The contradiction will thus be obtained if I can show that the $C_n$'s are nowhere dense (by Baire). How can I show this? Or is there another way to attempt this question?
 A: Counterexample. The closed unit interval $X=[0,1]$ is a complete metric smace. Let $B=X\cap\mathbb Q,$ the set of all rational numbers in $X;$ and let $A=X\setminus\mathbb Q,$ the set of all irrational numbers in $X.$ Thus $X=A\cup B;$ and $B,$ being a countable set, is a countable union of closed nowhere dense sets. The Cantor set $C$ is a closed nowhere dense set; I claim that $A$ is homeomorphically embeddable in $C.$
I. If you happen to know that $C$ is homeomorphic to $\{0,1\}^\mathbb N$ and $A$ is homeomorphic to $\mathbb N^\mathbb N,$ you can simply observe that there is an obvious embedding of $\mathbb N$ into $\{0,1\}^\mathbb N,$ and therefore there is an embedding of $\mathbb N^\mathbb N$ (which is homeomorphic to $A$) into $(\{0,1\}^\mathbb N)^\mathbb N,$ which is homeomorphic to $\{0,1\}^{\mathbb N\times\mathbb N},$ which is homeomorphic to $\{0,1\}^\mathbb N,$ which is homeomorphic to $C.$
II. Each number $x\in A=X\setminus\mathbb Q$ can be represented uniquely in the form
$$x=\sum_{n=1}^\infty\frac{\varepsilon_n}{2^n},\ \varepsilon_n\in\{0,1\};$$
and you can verify that the function
$$f(x)=\sum_{n=1}^\infty\frac{2\varepsilon_n}{3^n}$$
maps $A$ homeomorphically onto a subspace of $C.$
