Let $(X,d)$ be a metric space and $(x_n)\subseteq X$ . Then any sub sequential limit of $(x_n)$ is either a limit point of $\{$ $x_n$ $\}$ or an isolated point of $\{$ $x_n$ $\}$ and the limit of a constant subsequence.
My attempt is as follows:
Let $A= $ $\{$ $x_n$ $:$ $n\in \mathbb{N} $} $=$ $\{$ $x_n$ $\}$
Suppose $x$ is the limit of a subsequence $(x_{n_k})$. Suppose $x$ is not a limit point of $\{$ $x_n$ $\}$ . So there exists a radius $r>0$ such that $B(x,r)\cap A$ $\backslash $ $\{$ $x$ $\}$ $=$ $\varnothing$. If $x\in A$ then $x$ is an isolated point of $\{$ $x_n$ $\}$ . Since $x$ is the limit of $(x_{n_k})$, there exists infinitely many $(x_{n_k})$ in $B(x,r)$. Since the intersection of $B(x,r)$ and $A\backslash \{$ $x$ $\}$ is empty, it follows that $x_{n_k}$ must be $x$. If $x\notin A$ and $x$ is not a limit point of $A$ then there exists an open set $U$ containing $x$ such that $U\cap A= \varnothing$ . Since $x$ is the limit of a subsequence $(x_{n_k})$, it follows that there exists $K$ such that $k\geq K$ implies $x_{n_k} \in U$, a contradiction, unless $x_{n_k}=x$ for each $k$.
Is my proof correct?
In essence your proof will work, but maybe write it down more directly? (no $x \in A$ or $x \notin A$ distinction needed; it follows from the proof)
So assume $x$ is the limit of the subsequence $(x_{n_k})_{k \in \mathbb{N}}$ with $n_1 < n_2 < n_3 < \ldots$ and define $A = \{x_n: n \in \Bbb N\}$ the set of values.
If $x \notin A'$ there is some $r>0$ such that $B(x,r) \cap (A\setminus\{x\})=\emptyset$. For this $r>0$ we find some $N$ such that $k \ge N$ implies $x_{n_k} \in B(x,r)$, and this means, as $x_{n_k} \in A$, that $x_{n_k}=x$ for all $k \ge N$ (or else it would be in $B(x,r) \cap (A\setminus \{x\})$ which cannot be by the choice of $r$). Then this tail of the subsequence (which is also a subsequence) shows that $x$ is the limit of a constant subsequence (with value $x$ of course ) of the original sequence. As $B(x,r) \cap A=\{x\}$, this also shows that $x$ is an isolated point of $A$. So the dichotomy has been shown.