Closed subset of $\mathbb R^2$ disconnecting $\mathbb R^2$ must contain a curve? Edit: Per Alex S's answer, my first statement was false, so I've moved it to the bottom.
Here is the weaker statement that I am trying to prove (this was included in my original question): Let $C \subset \mathbb R^2$ be a closed subset such that $\mathbb R^2-C$ is disconnected. I suspect that $C$ should contain a connected component with more than one element. Edited addendum: As an intermediate between this question and my first one below, one could also ask whether $C$ has a path-connected component with more than one element.
Context: This the approach I wanted to take in answering this question.
Original question: Let $C \subset \mathbb R^2$ be a closed subset such that $\mathbb R^2-C$ is disconnected. I strongly suspect that $C$ should contain a curve $\gamma: \mathbb R \to C$ which also disconnects the plane, that is, $\mathbb R^2-\gamma(\mathbb R)$ is disconnected, but I don't know how to prove it. It seems to me like one should be able to try and "travel along a boundary" of $C$, but these kinds of things are always difficult to make precise in topology.
Note that $C$ is precisely the complement of two disjoint nonempty open subsets of $\mathbb R^2$; this perspective may be easier to approach conceptually. Also, if $C$ has nonempty interior, then this property is automatically satisfied - simply take a circle in the interior. As that example illustrates, I'm not requiring that the connected components of $\mathbb R^2 - \gamma(\mathbb R)$ "respect" any disconnecting of $\mathbb R^2-C$.
 A: Your primary claim is false, even you require that $C$ be compact. Consider the Warsaw curve:

It is the closure of the graph of $\sin(1/x)$ for $x\in[\pi,\infty)$, together with a simple arc joining $(1,0)$ and $(0,0)$. This separates the plane, but it does not contain a loop that also separates the plane. 
Of course, you are looking for the image of a map $\gamma:\mathbb R\to C$ to separate the plane, not just a loop. For that we have to change the example. Start with the closure of the graph of $\sin(1/x)$ for $x\in[\pi,\infty)$, and union that with the closure of the graph of $\sin(1/(x-1))$ for $x\in[\pi,\infty)$, and finally union with a simple arc from $(2,0)$ to $(0,0)$. We can avoid one of the problem points by having $\gamma(t)$ limit to, but never reach that point at $t\to\pm\infty$. But we cannot avoid both points.
At this time, I do not know about your weaker claim.
A: R. H. Bing constructed a counter example to your intermediate question, see Eric Wosey's answer here:  It is a totally path disconnected compact subset of $R^2$ which separates $R^2$ (a pseudocircle). However, a totally disconnected closed subset of $R^2$ cannot separate, see George Lowther's answer to the linked question.   
