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 $X$.