Prove or disprove that a Jordan curve exists that contains two points. Question (not homework): Let $\Omega$ be open and connected.  Prove or disprove that if $z_1, z_2\in\Omega$ are distinct, then there exists a Jordan curve $\gamma\subseteq\Omega$ such that $z_1, z_2\in\mbox{int }\gamma$.
Partial answer: I think that this statement is true.  Path-connectivity and connectivity are equivalent for open subsets of $\mathbb{C}$; thus there exists a path $\gamma_1$ connecting $z_1$ and $z_2$.  If there exists another path $\gamma_2$ disjoint from $\gamma_1$ then we are done: the path $\gamma_1\cup \gamma_2$ can be perturbed to go around $z_1$ and $z_2$ along an arc with small radius.
But how do we prove the existence of such a $\gamma_2$?  By openness of $\Omega$, for every $z\in\gamma_1$ there exists $\epsilon_z > 0$ such that $B(z, \epsilon_z)\subseteq \Omega$, and we can choose a point in the punctured ball (and by continuity, on the "same side" of the curve for each choice).  But such an algorithm requires the Axiom of Choice, and I'm all but certain we can do better and simpler.
Another approach is to consider $(\Omega - \gamma_1)\cup\{z_1, z_2\}$: if this set is connected, then the conclusion follows.  The question is how we know that it's connected: it "seems obvious," but of course we need a proof.
What am I missing, if anything?
 A: It is true, but the proof is technically complicated.
We shall prove that there exists a polygon Jordan curve $\gamma\subseteq\Omega$ such that $z_1, z_2\in\mbox{int }\gamma$.
A Jordan curve in $\Omega$ is a map $\gamma : [a,b] \to \Omega$ such that $\gamma(s) = \gamma(t)$ only for $\{s,t\} = \{a,b\}$. Often $\gamma$ is identified with its image $\gamma([a,b])$ which is a space homeomorphic to the unit circle $S^1$.
A polygon Jordan curve in $\Omega$ is a Jordan curve $\gamma : [a,b] \to  \Omega$ admitting a partition $t_0 = 0 < t_1 < \ldots < t_n = 1$ such that $\gamma \mid_{[t_{i-1},t_i]}$ is a linear parameterization of the geometric line segment $L(z_{i-1},z_i) = \{ z_{i-1} + t(z_i-z_{i-1}) \mid t \in [0,1]\}$ from $z_{i-1}$ to $z_i$, where $z_j = \gamma(t_j)$. The specific linear parameterizations of the $L_i$ are irrelevant for us, therefore instead of polygon curves $\gamma : [a,b] \to \Omega$ it suffices to consider tupels $\zeta = [z_0,\ldots,z_n]$ of points in $\Omega$ such that

*

*$z_n = z_0$.


*All geometric line segments $L_i = L(z_{i-1},z_i)$, $i=1,\ldots, n$, are contained in $\Omega$.


*$L_i \cap L_{i+1} = \{z_i\}$ and  $L_1 \cap L_n = \{z_0\}$, all other intersections of line segments $L_i, L_j$ with $i < j$ are empty.
See the answer to The necessary and sufficient conditions for a complex function to have primitive function for more details.
Given any tupel $\zeta$  as above, we call $J(\zeta) = \bigcup_{i=1}^n L_i$ a Jordan polygon. Note that we must have $n \ge 2$ to get a Jordan polygon. As arbitrary Jordan curves, Jordan polygons divide the plane into an interior $\operatorname{int} \zeta$ and an exterior $\operatorname{ext} \zeta$.
Let $B_r(x) \subset \mathbb C$ and $D_r(x) \subset \mathbb C$ denote the open and disk with center $x \in \mathbb R^n$ and radius $r > 0$, respectively.
Let $U \subset \Omega$ denote the set of points  $y \in \Omega$ such that $x_1, y$ are contained in the interior of some Jordan polygon. Clearly $x_1 \in U$ (take the boundary of a small square with center $x_1$). It is obvious that $U$ is open. We shall show that $U$ is also closed in $\Omega$; this implies $U = \Omega$ because $\Omega$ is connected. Hence there is a Jordan polygon containing $x_1, x_2$ in its interior.
Assume that $U$ is not closed in $\Omega$. Then there exists $z \in \Omega \setminus U$ which is a cluster point of $U$. Choose $r > 0$ such that $B_{r}(z) \subset \Omega$. There exists $y \in U \cap B_r(z)$. Choose a Jordan polygon $J(\zeta)$ containing $x_1,y$ in its interior. $J(\zeta)$ must intersect $B_r(z)$; otherwise we would have either $B_r(z) \subset \operatorname{int} \zeta$ (which implies $z \in \operatorname{int} \zeta$, i.e. $z \in U$, a contradiction) or $B_r(z) \subset \operatorname{ext} \zeta$ (which implies $y \notin \operatorname{int} \zeta$, a contradiction).
In the sequel it will be useful to draw pictures to see what is going on; I leave this to you.
Case 1. $z \in J(\zeta)$.
W.l.o.g. we may assume that $z = z_i$ for some $i$ (otherwise we insert $z$ between suitable consecutive $z_{i-1}, z_i$; the extended tupel gives the same Jordan polygon). We may moreover assume that $0 < i < n$ (otherwise we replace $\zeta = [z_0,\ldots,z_n]$ by $\zeta' = [z_1,\ldots,z_n,z_1]$ which gives the same Jordan polygon as $\zeta$). Choose any line $L$ through $z_i$ such that $L \cap L_i = L \cap L_{i+1} = \{z_i\}$. Choose $d > 0$ such that $d < r$ and $D_r(z_i) \cap L_j = \emptyset$ for $j \ne i, i+1$. For $j = i, i+1$ let $z'_j$ be the unique point of $L_j$ such that $\lvert z'_j - z_i \rvert = d$ and let $z^*$ be the unique point on $L$ such that $z^* \in \operatorname{ext} \zeta$ and $\lvert z^* - z_i \rvert = d$ (note that $L$ contains two points having distance $d$ from $z_i$; one of them lies in  $\operatorname{int} \zeta$, the other in $\operatorname{ext} \zeta$). Then $\zeta' = [z_0,\ldots,z_{i-1},z'_{i-1},z^*,z'_{i+1},z_{i+1},\ldots,z_n]$ gives a Jordan polygon such that $\operatorname{int} \zeta  \cup \{z_i\} \subset \operatorname{int} \zeta'$. This shows $z_i = z \in U$, a contradiction.
Case 2. $z \notin J(\zeta)$.
Let $d>0$ be the distance from $z$ to $J(\zeta)$; clearly $d < r$ since $J(\zeta)$ intersects $B_r(z)$. Let $z' \in J(\zeta)$ be a point such that $\lvert z - z' \rvert = d$. As in case 1 we may w.l.o.g. assume that $z' = z_i$ for some $i$ and that $0 < i < n$. Consider the ray $R(z,z_i)$ starting at $z$ and going through $z_i$. It is impossible that both $L_i$ and $L_{i+1}$ lie on $R(z,z_i)$. W.l.o.g. we may assume $L_i$ does no lie on $R(z,z_i)$. Choose a point $w$ between $z_{i-1}$ and $z_i$ such that $w \in B_r(z)$. Then $L(w,z) \subset B_r(z)$. The tupel $\zeta' = [z_0,\ldots,z_{i-1},w,z,z_i,\ldots,z_n]$ gives again a polygon Jordan curve which has the property that $\operatorname{int} \zeta \subset \operatorname{int} \zeta'$ and $z  \in J(\zeta')$. We are now in the situation of case 1 which shows $z \in U$, a contradiction.
