prove the space is Polish

Question

Suppose $X=A\cup B$ is a separable metrizable space and $X$ is Baire. If $A$ is an open Polish subspace of $X$, $B$ is countable and closed nowhere dense in $X$, $A\cap B=\emptyset$, then is $X$ a Polish space?

Answer 1

It seems that $X$ needs not be Polish.

Take any perfect and nowhere dense closed set $E\subset\mathbb R$. Let $B$ be a countable dense subset of $E$ and $A:=\mathbb R\setminus E$. Then $X:=A\cup B$ is a Baire space since it is comeager in $\mathbb R$. The set $A$ is open in $\mathbb R$, hence Polish and open in $X$. The set $B$ is countable and nowhere dense in $X$ because $A$ is dense in $\mathbb R$ and hence in $X$. But $X$ cannot be Polish since $B$ is closed in $X$ and not Polish (being countable without isolated points).

I hope this works!