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Let $X$ be an open , simply connected (path connected with trivial first homotopy group) subset of $\mathbb R^2$. Let $0\in X$.

Is it true that for every $p,q\in X\setminus \{0\}$, there is an open , simply connected subset $A_{pq}\subseteq X\setminus \{0\}$ such that $p,q\in A_{pq}$ ?

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Yes. In fact, the assumption that $X$ is simply connected is unnecessary (you just need it to be connected and open), so we may replace $X$ with $X\setminus\{0\}$ to simplify notation.

To prove this, note that if $p\neq q$, we can find a polygonal path from $p$ to $q$ in $X$ which does not intersect itself. Taking sufficiently small open balls around each segment of this polygonal path, we get a simply connected open subset of $X$ containing $p$ and $q$. (You just have to choose your balls so that the ball around each segment only intersects the balls around the two neighboring segments and no others. Then you can show by induction that the union of the first $n$ balls is simply connected, since the intersection of the $n$th ball with the union of the first $n-1$ balls is convex and in particular connected.)

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