Inserting an open and simply-connected set between a compact set and an open set In a paper  I am reading, the following is considered obvious:
Let $K$ be a compact and connected subset of $\,\mathbb R^2$, with $\mathbb R^2\smallsetminus K$ also connected, and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open  $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected $($and $K$ not necessarily connected$)$ and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open  $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.
I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries. 
Any ideas?
 A: The Warsaw circle is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement. This provides a counterexample for the question as originally worded but does not have connected complement.
It can be realized as the union of the following:
$$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$
$$0\times[-2,1]$$
$$[0,1]\times -2 $$
$$1\times[-2,\sin 1]$$

A: This is only a partial answer, however it is too long for a comment.
What you want is actually true for compact subsets $K \subset \mathbb{R}^n$, i.e. in any dimension, under the additional assumption that $K$ is locally contractible, i.e. that any point $x \in K$ has a contractible neighborhood.
This is a consequence of the following result, that can be found in Hatcher's book Algebraic Topology, Theorem A.7. page 525 in the Appendix.

Theorem. A compact subspace $K \subset \mathbb{R}^n$ is a retract of some (open) neighborhood $V$ if and only if it is locally contractible in the weak sense that for each $x \in K$ and each neighborhood $A$ of $x$ in $K$ there exists a neighborhood $B \subset A$ suct that the inclusion map $B \hookrightarrow A$ is nullhomotopic.

Now, if $K$ is retract of some neighborhood, it is also a retract of any smaller neighborhood, just by restriction of the retraction. So, if the assumption of the theorem is satisfied, you can find a simply connected open subset $V$ such that $K \subset V \subset U$.
This does not completely answer the question because there are compact, simply connected subspaces of $\mathbb{R}^n$ that are not locally contractible. An example in $\mathbb{R}^2$ is the comb space, which is compact, contractible (hence simply connected)
but not locally connected, hence not locally contractible.
An example in $\mathbb{R}^3$ is given by the cone over the Hawaiian earring, which is contractible but not even locally simply connected.  
A: Perhaps the theorem at the common intersection of the various ambiguous interpretations of the inquiry at hand is the following: 
If the planar open set U contains a continuum K and if the planar complement of K is connected, then there exists a connected and simply connected open set V such that K is a subset of V, and V is a subset of U.
This follows, for example, from the fact that K is cellular, the nested intersection of closed topological planar disks. Surround K by a slightly larger disk D, so that D is contained in U, and the interior of D is the coveted open set V.
For a barehanded argument of the cellularity claim, prove by induction that if C is a planar simple closed curve, and if C is the cancatanation of vertical and horizontal segments of length 1, then C bounds a topological disk. 
To obtain the mentioned disk D, tile the plane by very small squares, and extract a curve C from the boundaries of the tiled squares.
Of course the argument sketched above does not use any heavy artillery such as the Jordan curve theorem or the Schoenflies theorem.
A: Pick a point $x_i$ in every connected component $X_i$ of $\mathbb{R}^2 \backslash U$. Since $\mathbb{R}^2 \backslash K$ is connected and open, it is path connected. Fix some distinguished $x_0 \in \mathbb{R}^2 \backslash K$, and create a path $\gamma_i$ connecting $x_i$ to $x_0$ in $\mathbb{R}^2 \backslash K$. Let $V$ be $U$ minus the union of these paths. As long as there are a finite number of  $X_i$, $V$ will be open and all its connected components will be simply connected. So you just need to show that you can replace $U$ with a small open subset containing $K$ whose complement has finitely many connected components.. A proof eludes me right now.
