Let $\gamma:[0;1]\to\mathbb{R}^2$ be Jordan curve and $p,q\in\mathbb{R}^2\setminus\mathrm{Im}(\gamma)$. Suppose there exists $\alpha:[0;1]\to\mathbb{R}^2$ join p to q, $\alpha$ is not intersect with $\gamma$. Proof if $\mathrm{Im}(\alpha)\cap\mathrm{Im}(\gamma) =\emptyset$ then $p$ and $q$ are contained in a component of $\mathbb{R}^2\setminus\mathrm{Im}(\gamma)$.
Any hint will be appreciate.
Suppose not. Let $X = \Bbb R^2 \setminus \mathrm{Im}(\gamma)$.
Then since $p$ and $q$ are in different components of $X$, we know that there are open sets $U, V \subset X$ with $p \in U$ and $q \in V$ and $$ U \cap V = \emptyset\\ U \cup V = X.$$
Look at $U' = \alpha^{-1}(U)$, and $V' = \alpha^{-1}(V)$.
These are open sets in $[0, 1]$, with $0 \in U', 1\in V'$, and $$ U' \cup V' = [0, 1]\\ U' \cap V' = \emptyset. $$
They thus show that the unit interval is not a connected set; that's a contradiction.