Do we need to have a subsequence such that $\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|$? Let $(X,\left\|\;\cdot\;\right\|)$ be a normed space and $(x_n)_{n\in\mathbb{N}}\subseteq X$. Can we prove that there is a subsequence $\left(x_{n_k}\right)_{k\in\mathbb{N}}\subseteq(x_n)_{n\in\mathbb{N}}$ such that $$\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|\;?$$
 A: Your question can be greatly reduced. The normed space and its norm can be "forgotten" because both the limit and the $\liminf$ are applied to sequences of real numbers ($||x_n||$ is a real number!), and the question simply becomes:

For a given sequence of positive real numbers $(a_n)_{n\in \mathbb N}$, does there exist a subsequence $(a_{n_k})_{k\in\mathbb N}$ such that $$\lim_{k\to\infty} a_{n_k}= \liminf_{n\to\infty} a_n$$

This question is much simpler to answer and is almost trivial, depending on your definition of $\limsup$.
A: Yes, and this holds more generally for real sequences. In fact, the limit inferior of a sequence can be defined as the least limit point of any subsequence of the sequence.
To prove that such a subsequence exists, recall that,
$$\liminf\limits_{n\rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \inf\limits_{k \ge n} a_k$$
Let the limit inferior be $L$ and fix an integer $m  > 0$. By the above limit being $L$, we may choose some $n$ such that
$$\left| \inf\limits_{k \ge n} a_k - L\right| < \frac{1}{2m}.$$
By definition of an infimum, we may choose some $n_m \ge n$ such that,
$$\left| \inf\limits_{k \ge n} a_k - a_{n_m}\right| < \frac{1}{2m}.$$
Using the triangle inequality,
$$\left| a_{n_m} - L\right| \le \left| \inf\limits_{k \ge n} a_k - L\right| + \left| \inf\limits_{k \ge n} a_k - a_{n_m}\right| < \frac{1}{m}.$$
By squeeze theorem, $a_{n_m} \rightarrow L$.
A: Yes: this is the same as proving that for a real sequence $(a_n)_k$ there is a subsequence $(a_{n_k})_k$ such that
$$ \lim_{k} a_{n_k} = \liminf_n a_n = l, $$
say.
To do this, we basically imitate one of the proofs of Bolzano–Weierstrass (the "high points" one) (or see here): pick a sequence $\varepsilon_k \to 0$. Then for each $\varepsilon_k$, there are, by the definition of the limit inferior, an infinite number of $n$ such that $ \lvert a_{n} - l \rvert < \varepsilon_k $. Choose $n_k$ as the smallest of these $n$ not already chosen, and then you find that $(a_{n_k})_k$ satisfies $ 0 \leqslant \lvert a_{n_k} - l \rvert < \varepsilon_k \to 0$, so $a_{n_k} \to l $.
