Proving that $ \liminf_{n\to \infty} x_n=\min \mathcal{L}((x_n)_{n\ge 1})$

Question

For a bounded sequence of real numbers $(x_n)_{n\ge 1}$, I would like to prove that $\displaystyle \liminf_{n\to \infty} x_n=\min \mathcal{L}((x_n)_{n\ge 1})$, where by $\mathcal{L}((x_n)_{n\ge 1})$ I denoted the set of all limit points of $(x_n)_{n\ge 1}$.
My Proof: Consider $L\in \mathcal{L}((x_n)_{n\ge 1})$. We would like to prove that $\liminf\limits_{n\to \infty}x_n\le L$.
Since $L\in \mathcal{L}((x_n)_{n\ge 1})$, there will be a subsequence $(x_{k_n})_{n\ge 1}$ such that $L=\lim\limits_{n\to \infty} x_{k_n}$.
By taking into account the definition of limit inferior, our conclusion rewrites as $\displaystyle\lim_{n\to \infty}\inf_{k\ge n}x_k \le \lim_{n\to \infty}x_{k_n}$.
We are going to prove that $\displaystyle\inf_{k\ge n}x_k\le x_{k_n}, \forall n\in \mathbb{N}$. This is obviously true, since we have that $k_n\ge n$, $\forall n \in \mathbb{N}$, so it follows that $x_{k_n}\in \{x_k | k\ge n\}, \forall n \in \mathbb{N}$ and from here we have that $\displaystyle x_{k_n}\ge \inf_{k\ge n}x_k$, $\forall n \in \mathbb{N}$.
If we take the limit as $n\to \infty$ in our last equality, we get that $\displaystyle\lim_{n\to \infty}\inf_{k\ge n}x_k \le \lim_{n\to \infty}x_{k_n}$, which means that $\displaystyle \liminf_{n\to \infty}x_n \le L$.
Since $L$ was taken arbitrarily, our conlcusion follows.
I would like to know if my reasoning is correct. My textbook proves this result in a different way, but I think that my way is also correct.