Edited as suggested in the comments.
Prove that if $x$ is a cluster point of a real sequence $(x_n)_n$, it is a limit point of a subsequence of $(x_n)_n. $
Previous questions have already tackled this issue but I want to be sure that the specific wording in the proof I came up with is okay.
Let $(x_n)_{n \in \mathbb{N} } $ be a sequence of real numbers.
Assume $x$ is a cluster point of $(x_n)_n$.
Then, $\forall \epsilon >0 $ and $\forall n \in \mathbb{N}, \exists m>n $ such that $d(x,x_m)< \epsilon$.
Then we can construct a subsequence $(y_n)_n$ of $(x_n)_n$:
$y_1 \equiv x_{k_1}$ where $k_1 \equiv \min \{n>0: 0<d(x, x_n)<1 \} $
$y_2 \equiv x_{k_2}$ where $ k_2 \equiv \min\{ n>k_1 : 0<d(x,x_n)<d(x,x_{k_1}) \}$
for any $y_n$ with $n>2$, $y_n \equiv x_{k_n}$ where $k_n \equiv \min \{ n>k_{n-1}:0<d(x,x_n)<d(x,x_{k_{n-1}}) \}.$
Note that $k_n$ is well-defined since the definition of cluster point guarantees its existence.
Since by construction $\forall n,m \in \mathbb{N}, m>n, d(x,y_m)<d(x,y_n)$, it is the case that $d(x,y_n) \rightarrow 0 $ as $n \rightarrow0$.
Therefore $\forall \epsilon>0, \exists N \in \mathbb{N} $ such that $\forall n >N, d(x,y_n)< \epsilon$, i.e. $x$ is the limit point of the subsequence of $(x_n)_n$, $(y_n)_n$.
Thanks in advance!!