This is half of Theorem 3.19 from Baby Rudin. Rudin claims the proof is trivial. What I've come up with so far doesn't seem trivial, however, and is probably also wrong (my problem with it is pointed out below). This makes me wonder whether I'm overlooking some useful fact and/or using an unprofitable approach.
Theorem. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then
$$ \liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n. $$
Proof. Suppose that $s_n \leq t_n$ if $n \geq N$, but that
$$ \liminf_{n \rightarrow \infty} s_n > \liminf_{n \rightarrow \infty} t_n. $$
Let $E_s$ denote the set of all subsequential limits of $\{s_n\}$, and let $E_t$ denote the set of all subsequential limits of $\{t_n\}$. Then
$$ \inf E_s > \inf E_t. $$
This implies that there exists some $x \in E_t$ such that $\inf E_s > x > \inf E_t$, since otherwise $\inf E_t$ would not be the greatest lower bound of $E_t$. Hence some subsequence of $\{t_n\}$, say $\{t_{n_i}\}$, converges to $x < \inf E_s$.
Lemma. (from Rudin) If $x < \liminf_{n \rightarrow \infty} s_n$, then there exists an integer $N$ such that if $n \geq N$, then $s_n > x$.
By the lemma, there exists an integer $N_0$ such that if $n \geq N_0$, then $s_n > x$.
Now, let $\epsilon = \inf_{n \geq N_0} \{s_n – x\}$. This is where I think my proof breaks down. Can't $\epsilon$ be zero? Then, since $\{t_{n_i}\}$ converges to $x$, there exists an integer $N_1$ such that if $n_i \geq N_1$, then $|t_{n_i} – x| < \epsilon$. But this means that, if $n_i \geq \max\{N, N_0, N_1\}$, we have both
$$ s_{n_i} > t_{n_i}, $$
as well as $s_{n_i} \geq t_{n_i}$, a contradiction.
Best Answer
You are making it too hard for yourself.
By definition, $\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k\geq n} a_k$. Also, notice that the sequence $\inf_{k\geq n} a_k$ is non-decreasing.
So, if $s_n \leq t_n$, then clearly $\inf_{k\geq n} s_k \leq t_n$, for any $n$. From this it follows that $\inf_{k\geq n} s_k \leq \inf_{k\geq n} t_k$, for any $n$ (if not, then $\inf_{k\geq n} s_k > t_{k'}$, for some $k' \geq n$, which is an immediate contradiction).
Since both sides are non-decreasing, we have $\inf_{k\geq n} s_k \leq \lim_{n' \to \infty} \inf_{k\geq n'} t_k = \liminf_{n \to \infty} t_n$, and then we have $\lim_{n' \to \infty} \inf_{k\geq n'} s_k = \liminf_{n \to \infty} s_n \leq \liminf_{n \to \infty} t_n$, as desired.