Not sure if my proof as is is correct:
Since $a_n \leq b_n$ for all $n$, we have $a_n - b_n \leq 0$. As such, by the limit location theorem it holds that $\lim_{n\to\infty} (a_n - b_n) \leq 0$. By linearity, we have $\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \leq 0$, which implies $lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.
My one concern is using the LL Theorem on the sequence $a_n - b_n$: do I need to prove this sequence is convergent first?
On
Let $\lim_{n\to\infty} (a_n)=a$ and $\lim_{n\to\infty} (b_n)$=b ,then using epsilon definition of limit and this property of real numbers: $|a-b|\leq|a|+|b|$ you can prove easily that $\lim_{n\to\infty} (a_n-b_n)=a-b,$
Then in order to prove the inequality above that $\lim_{n\to\infty}(a_n)\leq\lim_{n\to\infty}(b_n)$ prove first that for every $c<a$ it holds $\neg \lim_{n\to\infty} (b_n)=c$
So it amounts to prove that if $u_{n}\rightarrow u$ and $u_{n}\leq 0$, then $u\leq 0$.
Assume that $u>0$, consider the positive number $u/2$, then $|u_{n}-u|<u/2$ for large $n$, and hence $u_{n}>u-u/2=u/2>0$, contradicts the assumption that $u_{n}\leq 0$.