I want to find a subset $D$ of the plane such that has two infinite setss $A,B\subseteq D$ for which every finite set $X\subseteq D$ there is an $A-B$ arc in $D-X$ but $D$ contains no infinite set of disjoint $A-B$ arcs.
This is a problem from Diestel's book of Graph theory, 5th edition in connectivity chapter. It seems analogous to Menger's theorem but I cannot imagine such that set $D$
Let $A$ be the set $\mathbb{N}\times {0} = \{(n,0) : n\in \mathbb{N}\}$ and $B = \{(-n,1) : n\in \mathbb{N}\}$.
To construct $D$, we will add a collection of arcs $P_n$, $n\in \mathbb{N}$ to $A\cup B$.
Let $P_0$ be the straight line from $(0,1)\in B$ to $(0,0)\in A$.
Let $P_n$ be the arc that starts at $(n,0)\in A$, follows a straight line to the point $(n, \frac{1}{2^n})$, then goes straight to the point $(-n, \frac{1}{2^n})$, then straight up to $(-n, 1)\in B$.
Now set $D = \bigcup\{P_n : n\in \mathbb{N}\}$.
It's clear that no finite subset $X$ separates $A$ from $B$, as no finite set contains a point of every arc $P_n$.
To show that there is no infinite set of disjoint $A-B$ arcs, consider a set $S$ of disjoint $A-B$ arcs. Let $Q$ be the arc in this set $S$ that has the leftmost starting point in $A$. Let $(a,0)\in A$ be the starting point of $Q$ and $(-b, 1)\in B$ the ending point. Every other arc in $S$ starts to the right of $(a,0)\in A$, and cannot cross $Q$. Thus every other arc ends at one of the $b$ points of $B$ to the right of $(-b,1)$, so $|S| \leq b$, which means $S$ is finite.
Here's a picture demonstrating the set $D$, and an example of a possible $A-B$ path (Note that not every $A-B$ path is necessarily one of the $P_n$).