Existence of Closed Curves around Bounded Components 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?
 A: 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 $(k, l) \in S$.
