A nice countable family of open subsets of $\mathbb{R}$ My question is the following (under $\mathtt{ZFC}$), where I assume that the topology of $\mathbb{R}$ is the usual one:

Is there a countable family $(U_n)_{n\in\mathbb{N}}$ of non-empty open subsets of $\mathbb{R}$ such that for every $X\subseteq \mathbb{N}$ which is infinite and co-infinite, there is a an $x\in\mathbb{R}$ with $\{n\in\mathbb{N}\mid x \in U_n\} = X$?

Why? Why not?
Ideas?
Thanks!
 A: Let $f: \mathcal{P}(\mathbb{N}) \to \mathbb{R}$ with $\displaystyle f(N) = \sum_{n=1}^\infty \frac{\left\{
\begin{array}{3c}
1, & \text{if} & n \in N\\
0, & \text{if} & n \not \in N\\
\end{array}
\right.}{2^n}$
So this is basically $0.\text{<is 1 in N?>}\text{<is 2 in N?>}\text{<is 3 in N?>}\ldots$ in binary.
Since the binary digits of the real number will only be exactly the same if the original subsets of the natural numbers were exactly the same, we can conclude that $f(A) = f(B) \Longrightarrow A = B$.
And now we simply have $\displaystyle U_n = \{ f(N) \mid N \subseteq \mathbb{N} \land N \not = \emptyset \land N \not = \mathbb{N} \land n \in N\}$. Since $\{ n \}$ is such a subset of $\mathbb{N}$, $U_n$ is never empty. Since $U_n$ completely covers the binary numbers in the interval $(0, 1)$, but does not include the endpoints 0 and 1 as we disallow $N = \emptyset$ and $N = \mathbb{N}$, it is an open set.
Since $X$ has to be infinite and co-finite, we know that $X \not = \emptyset$ and $X \not = \mathbb{N}$.
Now, for every $X \subseteq \mathbb{N}$, let $x = f(X)$, therefore:
\begin{equation}
\begin{array}{*2c}
  & \{ n \in \mathbb{N} \mid x \in U_n \} \\
= & \{ n \in \mathbb{N} \mid \exists N \subseteq \mathbb{N} (x = f(N) \land N \not = \emptyset \land N \not = \mathbb{N} \land n \in N) \} \\
= & \{ n \in \mathbb{N} \mid \exists N \subseteq \mathbb{N} (f(X) = f(N) \land N \not = \emptyset \land N \not = \mathbb{N} \land n \in N) \} \\
= & \{ n \in \mathbb{N} \mid \exists N \subseteq \mathbb{N} (X = N \land N \not = \emptyset \land N \not = \mathbb{N} \land n \in N) \} \\
= & \{ n \in \mathbb{N} \mid n \in X \} \\
= & X \\
\end{array}
\end{equation}
A: Such a family exists:
take any Cantor scheme $(U_s)_{s\in 2^{<\mathbb{N}}}$ of open subsets of $\mathbb{R}$ and let $\varphi: 2^\mathbb{N}\rightarrow \mathbb{R}$ be the induced embedding.
Define $U_n$ for each $n\in\mathbb{N}$ as $U_n = \bigcup_{s\in 2^n} U_{s^\smallfrown 1}$, where here I consider $2^n$ to be the set of  finite binary sequences of length $n$, with $2^0$ being the singleton of the empty-sequence.
Now given any $A\subseteq\mathbb{N}$ let $\unicode{x1D7D9}_A \in 2^{\mathbb{N}}$ be the characteristic function of $A$, we have that $$\{n\in\mathbb{N}\mid \varphi(\unicode{x1D7D9}_A)\in U_n\} = A$$
