An arc is a homeomorphic copy of $[0,1]$.
Question 1. Let $U$ and $V$ be disjoint open subsets of $[0,1]^2$. Is there an arc $A\subseteq [0,1]^2$ such that $[0,1]^2\setminus A$ is the union of two disjoint open sets $T$ and $W$ such that $U\subseteq T$ and $V\subseteq W$?
I believe this is true but have not found a reference or proof.
The following question is probably equivalent:
Question 2. If $U$ and $V$ are disjoint open sets in the sphere $S^2$, then there is a simple closed curve $A$ such that $S^2\setminus A$ has two components, one containing $U$ and the other containing $V$.
Question 1: the closure of the graph of the topologist's sine curve (i.e., the function $x \mapsto \sin \frac{1}{x}$) over the interval $[0, \pi/2]$ is a closed set $C$ whose complement in the rectangle $R = [0, \pi/2]\times [-1, 1]$ is the union of two open sets $U$ and $V$. But for any arc $A \subseteq C$, $R \mathop{\backslash} A$ is is connected and hence not the disjoint union of two non-empty open sets. $R$ is homeomorphic to the square $[0, 1]^2$ and the properties you are interested in are preserved under homeomorphisms, so your claim does not hold.
Question 2: Consider $S^2$ as the quotient space obtained from $[-\pi/2, \pi/2]\times [-1, 1]$ by identifying the boundary to a single point and then consider the image of the topologist's sine curve over $[-\pi/2, \pi/2]$. Again your claim does not hold.