Existence of open set in $R^2$ such that all open sets in $R$ are contained among its vertical sections I'm self-studying set theory, and one of the exercises on building Borel $\sigma$-algebra $\mathcal{B}$ via transfinite recursion from $F_\sigma = \{ \text{all countable unions of all closed sets in } \mathbb{R} \}$ and $G_\delta = \{ \text{all countable intersections of all open sets in } \mathbb{R} \}$ asks proving the above claim.
I probably understand why such a set exists if we omit the requirement for it to be open: it's easy to show that $\big|\mathcal{B}\big| = \mathfrak{c}$, where $\mathfrak{c}$ is the cardinality of continuum, so for the set of open sets $\mathcal{O} \subset \mathcal{B}$: $\mathfrak{c} = \big|\{ (0; x) \mid x \in (0; \infty) \}\big| \leq \big|\mathcal{O}\big| \leq \mathfrak{c} \Rightarrow \big|\mathcal{O}\big| = \mathfrak{c}$, so we could just map $\mathcal{O}$ onto $\mathbb{R}$ and call it a day.
But, indeed, how does one deal with the set being open?
 A: Let $I_1,I_2,\dots$ be a list of all the open intervals with rational endpoints. Every open subset of $\mathbb R$ is the union of some of these. Furthermore (for technical convenience), arrange the listing so that each rational interval appears infinitely often in the list.  Also, let $J_n$ be the subset of $[0,1]$ consisting of those numbers whose binary expansion has a 1 in the $n$-th place. For those numbers with two binary expansions (the dyadic rationals), put them into $J_n$ iff both of the expansions have a 1 in the $n$-th place. So $J_n$ is an open set, the union of $2^{n-1}$ open intervals each having length $2^{-n}$.  I claim that the set $U=\bigcup_n(J_n\times I_n)$ almost works (and I'll correct the "almost" later).
Given any open set $G\subseteq \mathbb R$, let $A$ be the set of those $n$ such that $I_n\subseteq G$, and note that $G=\bigcup_{n\in A}I_n$. I need an $x\in\mathbb R$ such that the vertical section of $U$ at $x$ is $G$, and for this it (more than) suffices to get an $x$ such that
$$
x\in J_n\iff n\in A.
$$
This essentially tells me how to find $x$: its binary expansion should have 1 in the positions corresponding to numbers $n\in A$ and 0 in the other positions. 
There's just one problem, resulting in the "almost" earlier. What if that $x$ is a dyadic rational number and therefore isn't in all the $J_n$ that I want it to be in? Well, this is where I use that each rational interval occurs infinitely often in the list of $I_n$'s. So, if some bit in the binary expansion of $x$ is 1 (because the corresponding $I_n$ is a subset of $G$), then infinitely many bits in the binary expansion of $x$ are 1; and similarly for 0.  So if $x$ is a dyadic rational, and so its binary expansion has either only finitely many 0's or only finitely many 1's, then in fact it has either no 0's at all or no 1's at all. In other words, $A$ contains none or all of the natural numbers, and so $G$ is $\varnothing$ or all of $\mathbb R$. 
So what I have, so far, is that all open sets except $\varnothing$ and $\mathbb R$ occur as vertical sections of $U$. Of course, $\varnothing$ actually does occur, just by taking any $x$ outside $[0,1]$. And we can get $\mathbb R$ to occur by adding a vertical strip to $U$. For example, $U\cup((1,2)\times\mathbb R)$ has all open subsets of $\mathbb R$ as vertical sections.
A: First, an observation. Let $X$, $Y$ be topological spaces, $U$ open subset of $X\times Y$ such that every open subset of $Y$ is a section of $U$. Let $X\subset X_1$, $Y\supset Y_1$. Then we can produce an open subset for $X_1\times Y_1$ with the same property. Indeed, extend $U$ to $X_1\times Y$, then restrict to $X_1\times Y_1$. 
It's enough to produce an open subset of $I\times I$ with the property. 
 We use the fact that  there exists a continuous surjection from $[0,1]$ to $[0,1]^{\mathbb{N}}$  $t\mapsto (\psi_{n}(t))_n$ (see Theorem Hahn-Mazurkiewicz --link). Let the open subset $U$ of $I\times I$, $$U=\{(t,u)\ | \psi_{2n}(t)< u < \psi_{2n+1}(t) \text{ for some } n \}$$
Since every open subset of $I$ is a countable union of intervals, every open subset of $I$ will be a section of $U$. 
Another method, based on the idea of @Andreas Blass: . Let $Y$ a topological space with a countable basis $V_n$. Let $C$ be the Cantor set.  Consider $U$ the open subset of $C\times I$, $(x,y)\in U$ iff $x_n=1$ and $y \in V_n$ for some $n$. Now extend $U$ to an open subset of $\mathbb{R}\times Y$.
