proving that the if $s_{n} \leq t_{n}$ for $n\geq N$ $\liminf_{n\to \infty} s_{n} \leq \liminf_{n \to \infty}t_{n}$ Using Danny Pak-Keung Chan's answer I've tried to do the supremum case.
We want to show if $s_{n} \leq t_{n}$ then $\limsup s_{n} \leq \limsup t_{n}$
Let $n\geq N$ be arbitrary and let $k \geq N$ be arbitrary. Then $s_{k} \leq t_{k} \leq \sup_{m \geq n}t_{m}$. Thus $\sup_{m \geq n}t_{m}$ is an upper bound of the set $\{s_{n}, s_{n+1}, \dots ,\}$ therefore $\sup_{m \geq n}t_{m} \geq \sup_{m \geq n}s_{m}$.
Define $S_{n}:= \sup_{m \geq n}s_{m}$ and $T_{n}:= \sup_{m \geq n}t_{m}$. Then we have $S_{n}$ and $T_{n}$ are decreasing. Hence $S_{n} \leq T_{n}$. Therefore
$\lim_{n \to \infty}S_{n} \leq \lim_{n \to \infty}T_{n}$. Hence $\limsup s_{n} \leq \limsup t_{n}$
 A: Let $n\geq N$ be arbitrary. Let $k\geq n$ be arbitrary. We have
that $\inf_{m\geq n}s_{m}\leq s_{k}\leq t_{k}$. Since $k$ is arbitrary,
we conclude that $\inf_{m\geq n}s_{m}$ is a lower bound of the set $\{t_{n},t_{n+1},\ldots\}$.
Therefore, $\inf_{m\geq n}s_{m}$ is smaller than or equal to the
greatest lower bound of $\{t_{n},t_{n+1},\ldots\}$, i.e, $\inf_{m\geq n}s_{m}\leq\inf_{m\geq n}t_{m}$.
Denote $S_{n}=\inf_{m\geq n}s_{m}$ and $T_{n}=\inf_{m\geq n}t_{m}$.
Note that $(S_{n})_{n\geq N}$ and $(T_{n})_{n\geq N}$ are increasing and $S_{n}\leq T_{n}$,
so $\lim_{n}S_{n}\leq\lim_{n}T_{n}$. Hence, $\liminf s_{n}\leq\liminf t_{n}$.
The proof for limsup is similar.
A: Using Danny Pak-Keung Chan's answer I've tried to do the supremum case.
We want to show if $s_{n} \leq t_{n}$ then $\limsup s_{n} \leq \limsup t_{n}$
Let $n\geq N$ be arbitrary and let $k \geq N$ be arbitrary. Then $s_{k} \leq t_{k} \leq \sup_{m \geq n}t_{m}$. Thus $\sup_{m \geq n}t_{m}$ is an upper bound of the set $\{s_{n}, s_{n+1}, \dots ,\}$ therefore $\sup_{m \geq n}t_{m} \geq \sup_{m \geq n}s_{m}$.
Define $S_{n}:= \sup_{m \geq n}s_{m}$ and $T_{n}:= \sup_{m \geq n}t_{m}$. Then we have $S_{n}$ and $T_{n}$ are decreasing. Hence $S_{n} \leq T_{n}$. Therefore
$\lim_{n \to \infty}S_{n} \leq \lim_{n \to \infty}T_{n}$. Hence $\limsup s_{n} \leq \limsup t_{n}$
