I am stuck on part of a complex analysis proof that I think needs more justification than given. It's pretty purely a topological statement, but it may be that complex-analytic techniques would be useful. Basically, the statement boils down to this:
Let $U \subset \mathbb{C}$ be open and connected, and let $K \subset \mathbb{C} \setminus U$ be a bounded component of the complement. Then there exists a curve in $U$ bounding $K$.
Is there any way to prove this statement without the Jordan-Schoenflies lemma, i.e. in a more elementary manner? I have tried a lot of random stuff but nothing seems to work. In particular, if there is a way to prove it with the argument principle that would be ideal, but not necessary.
I appreciate any help at all.
EDIT: So I really think this is a worthwhile question so I put the pitiful bounty I could afford haha.
The things that need to be worried about are things like the complement of a Cantor set, and other lower-dimensional components of the complement.
So we consider the set
$U_1 = \lbrace z$ $|$ $\exists \text{ (simple) closed curve } \gamma \subset U \text{with } z \text{ in a bounded component of } \mathbb{C} \setminus \gamma \rbrace$
Then $U_1$ is open, since we can write it as the union of the (open) interiors of bounded components of all curves in $U$, say
$U_1 = \cup_{\gamma_{\alpha} \in \Gamma} (\cup_1^{n_{\alpha}} C_{\alpha, k})$ for $\Gamma$ the set of all closed curves in $U$.
Now each $\bar{C_{\alpha, k}}$ intersects $\gamma_{\alpha} \subset U$ so that $\bar{C_{\alpha, k}} \cap U \neq \varnothing$, and writing $U_1$ as a union of all such sets, which each have (connected) $U$ in common, we get that $U_1$ is connected. Since it's open and $\mathbb{C}$ is locally path-connected, $U_1$ is path-connected.
Now, we can iterate this construction, then; if $\Gamma_k$ is the collection of all closed curves in $U_k$, write
$U_{k+1} = \lbrace z$ $|$ $\exists \text{ (simple) closed curve } \gamma \in \Gamma_k \subset U_k \text{with } z \text{ in a bounded component of } \mathbb{C} \setminus \gamma \rbrace$
Then let $V = \cup U_k$. Then any closed curve in $V$ is compact, and since each $U_k$ is open it is thus contained in finitely many such. But this is a nested sequence, so it lies entirely within some $U_n$. But all points in the bounded components of the curve are in $U_{n+1} \subset V$ so that the curve is homologous to zero in $V$.
By a theorem, $V$ is thus simply connected. So the crux is to show that, in fact, $V = U_1$.
But that's where I get stuck. We can also use the fact that a set is simply connected iff the complement in the sphere is the (connected) component of infinity. To show that it contains this set is the difficult direction; that the complement of the infinity component contains $V$ is trivial.
A proof should not use any theorem with the name "Jordan" in it, nor the word "homotopy". These proofs are obvious. I would say that from dimension theory, the Painleve Theorem is ok to use, but otherwise to avoid dimension. I would also consider it a solution to show that the statement implies either the Jordan-Schoenflies theorem for dimension 2, or the Annulus Theorem for dimension 2. And if anyone has enough reputation, could they add Continuum Theory to the list of tags?
If $\mathbb{C} \setminus U$ has only finitely many components then you can proceed roughly as follows:
Choose $\varepsilon > 0$ less than the distance of $K$ to any other component of $\mathbb{C} \setminus U$. Then consider the set of all squares $$ Q_{k, l} = \{ x+iy \mid (k-1)\varepsilon \le x \le k\varepsilon, (l-1)\varepsilon \le y \le l\varepsilon \} \, $$ which have at least one point in common with $K$: $$ S = \{ (k, l) \in \mathbb Z \times \mathbb Z \mid Q_{k, l} \cap K \ne \emptyset \} \, . $$ $S$ has only finitely many elements. For each $(k, l) \in S$, define $\gamma_{k, l}$ as the "boundary path" of $Q_{k, l}$ in counter-clockwise direction. Each $\gamma_{k, l}$ consists of four straight line paths $\gamma_{k, l}^{(n)}$, $n = 1,2,3,4$. If such a line path has at least one point in common with $K$ then there exists (exactly) one other line path of a $\gamma_{k', l'}^{(n')}$ which is the inverse path of $\gamma_{k, l}^{(n)}$.
Finally, define $\gamma$ as the formal sum of all line path $\gamma_{k, l}^{(n)}$ which lie completely in $U$. This is a closed curve.
If $a \in K$ lies in the interior of a $Q_{K,L}$ then the winding number satisfies $$ \frac{1}{2 \pi i} \int_{\gamma}\frac{dz}{z-a} = \sum_{(k, l) \in S} \frac{1}{2 \pi i} \int_{\gamma_{(k, l)}}\frac{dz}{z-a} = \frac{1}{2 \pi i} \int_{\gamma_{(K, L)}}\frac{dz}{z-a} = 1 $$ because all contributions of line paths intersecting $K$ cancel each other. By continuity this holds for all $a \in K$.
If $a$ lies in a different component of $\mathbb{C} \setminus U$, then $$ \frac{1}{2 \pi i} \int_{\gamma}\frac{dz}{z-a} = \sum_{(k, l) \in S} \frac{1}{2 \pi i} \int_{\gamma_{(k, l)}}\frac{dz}{z-a} = 0 $$ because the integral vanishes for all $(k, l) \in S$.