= Solution
For any function $f:X\to\mathcal P(X)$, form the diagonal subset
$$
D=\{x\in X:x\notin f(x)\}.
$$
If $f$ were <surjective>, some $d\in X$ would satisfy $f(d)=D$. Then $d\in D$ would hold exactly when $d\notin D$, which is impossible. Thus \b[no set surjects onto its power set], including the empty-set case. This is <Cantor's theorem> proved by the <Cantor diagonal argument>.
Use <real-number pairing by separated digits> for an explicit <injection> $\mathbb R^2\to\mathbb R$. First map a real number to $(0,1)$ by $q(x)=\tfrac12+\pi^{-1}\arctan x$, and use its unique <binary expansion> that is not eventually all ones. If $q(x)=\sum_{n\geq1}a_n2^{-n}$ and $q(y)=\sum_{n\geq1}b_n2^{-n}$, define
$$
\boxed{J(x,y)=\sum_{n\geq1}\left(\frac{a_n}{3^{2n-1}}+\frac{b_n}{3^{2n}}\right).}
$$
This places the two binary digit strings in alternating ternary positions. Such ternary strings, using only digits zero and one, are unique: at a first differing position $k$, the difference $3^{-k}$ exceeds the maximum tail $\sum_{j>k}3^{-j}=\tfrac12 3^{-k}$. Thus the output recovers both digit strings and both inputs. The use of base three avoids any endpoint ambiguity caused by interleaving digits in base two.
For a fixed $A\subseteq\mathbb R^2$, the <affine sections of a planar set> are indexed by
$$
\mathcal I=\{(\mathbf a,\mathbf b)\in\mathbb R^2\times\mathbb R^2:\mathbf b\ne0\}.
$$
There is an <injection> $\iota:\mathcal I\to\mathbb R$, for example by applying $J$ to each coordinate pair and then once more to those two results. Suppose every subset of $\mathbb R$ were a section of $A$. Define a function from $\mathbb R$ to $\mathcal P(\mathbb R)$ by assigning to $\iota(\mathbf a,\mathbf b)$ the corresponding section, and assigning the empty set to every real outside $\iota(\mathcal I)$. <Injectivity> of $\iota$ makes this definition unambiguous. It would be a <surjection> onto $\mathcal P(\mathbb R)$, contradicting the first argument. \b[No planar set has every subset of the real line as a section.] Allowing different parametrizations of the same line does not evade this obstruction.
For all <countable sets>, however, \b[such a planar set does exist]. To build one, first encode every real sequence $s=(s_1,s_2,\ldots)$ by one real number. Let $a_{nk}$ be the canonical binary digits of $q(s_n)$, and choose the explicit indexing
$$
j(n,k)=\frac{(n+k-2)(n+k-1)}2+n\quad(n,k\geq1).
$$
This is a <bijection> from positive integer pairs to positive integers. Define the <coding of real sequences by ternary digits>
$$
c(s)=\sum_{n,k\geq1}\frac{a_{nk}}{3^{j(n,k)}}.
$$
The same first-differing-digit argument proves that $c$ is <injective>; all its values lie in $[0,1/2]$. Now take
$$
\boxed{A=\{(c(s),s_n):s\in\mathbb R^{\mathbb N},\ n\geq1\}.}
$$
For every nonempty <countable set> $S\subseteq\mathbb R$, choose an enumeration $s$ of $S$, repeating terms if $S$ is finite. On the vertical line with $\mathbf a=(c(s),0)$ and $\mathbf b=(0,1)$, the section of $A$ is precisely $S$: injective sequence coding means no other sequence can contribute extra values in this column. The empty set is the section on the line $x=2$, since all code columns have first coordinate at most $1/2$. This constructs the required <universal planar set for countable real sections>. No assertion about its other, nonvertical sections is needed.
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