$\mathbf {The \ Problem \ is}:$ Let $(\mathbb R^n, \tau)$ denotes the standard topology on $\mathbb R^n$ . Does there exist any compact subspace $S$ , not totally disconnected, and any $n \in \mathbb N(\gt 1)$ such that the number of path-connected components of $S$ is exactly $\aleph_0$ ??? Also, same problem for number of path-connected components being $2^{\aleph_0} .$
$\mathbf {My \ approach} :$ As the standard topology of $\mathbb R^n$ is metrizable, hence $S$ must be closed and $\mathbf {bounded}$ . But, I am confused how to produce an example of a bounded set to have number of path-connected components $\aleph_0$ or $2^{\aleph_0} .$
Please provide me a small hint about either a counter-example, if exists or how to disprove .
The set $\{0\}\cup\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$ has countably many connected components. And the Cantor set has uncountably many. Finally, both of them are compact.