I want to prove that if $f:\mathbb R \rightarrow \mathbb R$ is continuous and $x_n$ a bounded sequence, then $\liminf_{n\rightarrow \infty}f(x_n) < f(\liminf_{n\rightarrow \infty}x_n)$.
Suppose $\liminf_{n\rightarrow \infty}x_n=a$
$$f\text{ continuous }\Rightarrow \forall \varepsilon>0,\ \exists \delta>0,\quad |x_n - a|<\delta \Rightarrow |f(x_n)-f(a)|<\varepsilon. \tag{1}$$
It follows from the definition of limit inferior that there is a sequence $x_{n_k}$ for which $x_{n_k}\rightarrow a$. In other words, $\exists k_0 \in \mathbb N$, such that $\forall k>k_0$, $|x_{n_k}-a|<\delta$. Now, by (1) we get $|x_{n_k} -a|<\delta \Rightarrow |f(x_{n_k} - f(a)|<\varepsilon$, $\forall k>k_0$.
My question is: May I infer from all this that $\liminf_{n\rightarrow \infty}f(x_n) \leq f(\liminf_{n\rightarrow \infty}x_n)$? Why?
Let $a = \liminf_n x_n$, then there is some subsequence $x_{n_k} \to a$. Since $f$ is continuous, $f(x_{n_k}) \to f(a)$. Hence $\liminf_n f(x_n) \le \liminf_k f(x_{n_k}) = \lim_k f(x_{n_k}) = f(a)$.
Note: The only tricky part here is showing $\liminf_n f(x_n) \le \liminf_k f(x_{n_k})$. Let $I = \{ n_k \}$. Then we have $\inf_{k \ge n} f(x_k) \le \inf_{k \ge n, k \in I} f(x_k)$, and since both sides are non-decreasing, we have $\inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k)$ followed by $\lim_n \inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k)$. Hence
$$\liminf_n f(x_n) = \lim_n \inf_{k \ge n} f(x_k) \le \lim_n \inf_{k \ge n, k \in I} f(x_k) = \liminf_k f(x_{n_k})$$