Extracting monotone distinct sequence

Question

Let $(x_n)$ be a sequence in $\mathbb{R}$ and $x_n\rightarrow p$ for some $p\in \mathbb{R}$. Suppose $x_n=p$ for finitely many index. Then there is a strictly monotone distinct (every element is distinct) subsequence $(x_{n_k})$ (of course $x_{n_k}\rightarrow p$).
Please comment on the correctness of my following contruction. Thanks in advance.

I try to construct $(x_{n_k})$ in the following way
Clearly $(-\infty,p)$or $(p,\infty)$ contains infinitely many $(x_n)$.
Case 1: $(-\infty,p)$ contains infinitely many $(x_n)$. In this case, we can construct a strictly increasing subsequence. Choose any point from $(\infty,p)$ as $x_{n_1}$. For $k\ge 2$, note that there are infinitely many $n_k$ such that $x_{n_k} \gt \max \{x_{n_1}, \cdots ,x_{n_{k-1}}\}$. Such $n_k$ always exists as sequence element equals to $x_{n_1}, \cdots ,x_{n_{k-1}}$ are only finitely many and the fact $x_n\rightarrow p$. Choose $n_k$ such that $n_k \gt n_{k-1}$. I think $(x_{n_k})$ is the required subsequence.

Case 2:$(-\infty,p)$ contains infinitely many $(x_n)$. By the same way, we can have a strictly decreasing subsequence $(x_{n_k})$ with the required property.

Remark: If both $(-\infty,p)$ and $(p,\infty)$ contains infinitely many $(x_n)$, then we get both strictly increasing and decreasing subsequence with the required property.

Answer 1

Your proof is correct.

Implicitly, you are using the fact that a sequence converges to a limit if and only if for every open interval containing the limit, there are only finitely many values of the sequence outside of that open interval.