Let $K\subset\overline{\Bbb C}=\Bbb C\cup\{\infty\}$ be a continua, that is non empty, connected and compact subset. I am reading a paper in which it is stated that there exists a sequence $\{a_n\}_{n\ge1}\subset\Bbb C$ such that $K$ is its cluster set. Thus, I think, fixed an arbitrary point $z_0\in K\cap\Bbb C$, we should find a subsequence $\{n_k\}_k$ such that $a_{n_k}\to z_0$.
But, why is this true?
The space $X=\overline {\Bbb C}$ is metrizable and separable because it is homeomorphic to $S^2.$ Any subspace of a separable metrizable space is separable. So there exists a countable $S\subset K$ such that $Cl_K(S)=K.$
Now $K$ is compact and $X$ is Hausdorff, so $K$ is closed in $X.$ So for any $T\subset K$ we have $Cl_K(T)=Cl_X(T).$ In particular, $Cl_X(S)=Cl_K(S)=K.$
If $K$ has more than $1$ member then any $x\in K$ belongs to $Cl_X(S\setminus \{x\})=Cl_K(S\setminus \{x\}),$ otherwise $K=Cl_K(S)=Cl_K(\{x\})\cup Cl_K(S\setminus \{x\})=\{x\}\cup Cl_K(S\setminus \{x\})$ is the union of two disjoint non-empty sets that are closed in $K,$ making $K$ disconnected.
So if $K$ has more than $1$ point then for every $x\in K$ there is a sequence of members of $S\setminus \{x\}$ that converges to $x.$