I have a question regarding excercise A.4.5 from the book Infinite-dimensional topology of function spaces by J. van Mill.
It asks to prove that for a metric separable spase $X$, any closed subset $A$ in it and any its open neighbourhood $V$ and any countable collection $\mathcal{U}$ of $X$ there exists an open set $W$ such that $A\subset W$ and $\overline{W}\subset V$ with $\overline{W\cap U} = \overline{W}\cap\overline{U}$ for all $U\in \cal{U}$.
I am struggling with the solution, and I don't understand the one presented in the answers section (initially I came up with the same one, but noe I don't see how it works).
The solution relies on the previous excercise A.4.5 which sates that if $f:X\to [0,1]$ is a continuous function and $U$ is any subset of $X$, then $\overline{f^{-1}([0,\delta))\cap U} \neq \overline{f^{-1}([0,\delta)}\cap\overline{U}$ only for countably many $\delta$.
So, in our case, we pick a function $f:X\to [0,1]$ with $f(A) \subset \{0\}$ and $f(X\backslash V)\subset \{1\}$ and after that attempt to apply A.4.5 for every $U\in\mathcal{U}$ but as far as I can see, this works only in case the set $\mathcal{W}=\{f^{-1}([0,\delta)) : \delta\in [0,1]\}$ is uncountable. In case $\mathcal{W}$ is countable, we can't get for any $U\in\mathcal{U}$ an uncountable subset of $\mathcal{W}$ with the property that $\overline{f^{-1}([0,\delta))\cap U} \neq \overline{f^{-1}([0,\delta)}\cap\overline{U}$ for each its element.
Please help me out, is there something that I am missing in the solution, or there indeed is a gap in there?