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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 $n\ge 2$.

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 $K=D\cup M$.

Not always.

Let $K$ be a subset of an ambient space $V$ ($V=\mathbf{R}^2$ is fine, but doesn't matter) that is the closure of a discrete subset $D$, such that $K-D$ is homeomorphic to a segment. This exists in $\mathbf{R}^n$ for $n\ge 2$.

Then every closed subset of $V$ 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 $V$ 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.

[Edit: I initially described $K$ as subset of the sine curve, but this doesn't matter and complicates the description.]

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 $K=D\cup M$.

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 $n\ge 2$.

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 $n\ge 2$.

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 $K=D\cup M$.

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 discrete dense subset therein (take an isometric parametrization of this curve $\mathbf{R}_{\ge 0}\to\mathbf{R}^2$ and take the image of $\{\sqrt{n}:n\in\mathbf{N}\}$). Get a space $X$.

Then every closed subset that meets every component has to contain all this discrete set of points, and hence contains its closure, and hence contains $S$. But in a segment in a space homeomorphic to a circle, all points that at in the intrinsic interior of the segment are also in the interior of the circle within the segment. But here this circle should contain all this discrete cloud which accumulates on all of $S$. Hence no subset of $\mathbf{R}^2$ (or any larger space, such as $\mathbf{R}^n$, $n\ge 2$) homeomorphic to a circle can meet every component of $X$.

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 $n\ge 2$.