Not always.

Start from the usual $\sin(1/x)$ example, consider the segment $S$ joining $(0,\pm 1)$, and replace the curve $\sin(1/x)$, $0<x\le 1$ with a more and more dense discrete subset $D$ therein (take an isometric parametrization of this curve $\mathbf{R}_{\ge 0}\to\mathbf{R}^2$ and take the image of $\{\sqrt{n}:n\in\mathbf{N}\}$), so that the closure $K$ of $D$ equals $D\cup S$.

Then every closed subset of $\mathbf{R}^2$ that meets every component of $K$ has to contain all $D$, and hence contains its closure, and hence contains $S$. But if $j:[0,1]\to C$ is an injection of a segment in a circle, the interior of $j([0,1])$ in $C$ is equal to exactly $j(\mathopen]0,1\mathclose[)$; in particular, $j([0,1])$ can't have empty interior in $C$.

But if $C$ were a circle within $\mathbf{R}^2$ meeting every connected component of $K$, we would have $D\subset C$, hence $K\subset C$. Since $j(S)=S$ has empty interior in $K$ and $K\subset C$, it has empty interior in $C$. This is a contradiction with the above fact. 

The argument applies in every space containing a copy of $K$, and in particular in $\mathbf{R}^n$ for every $n\ge 2$.

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Minor variant: let $M$ be any compact subset with empty interior, which is not homeomorphic to any subset of a circle (e.g., the whole sine example in the plane, a sphere in a higher space). Let $D$ a discrete subset of $\mathbf{R}^n$ whose set of accumulation points is exactly $M$ (this exists). Then no subset of $\mathbf{R}^n$ homeomorphic to $C$ meets every component of the compact subset $K=D\cup M$.