Show that all elements of one sequence are less than all elements of another sequence. Let $\{a_n\}_1^\infty$ and $\{b_n\}_1^\infty$ be two sequences in $\mathbb{R}$ such that $\forall n \in \mathbb{N}$, it is true that $a_n \leq b_n, a_n \leq a_{n+1}, \text{ and } b_{n+1} \leq b_n$.
We want to show $\forall m,n \in \mathbb{N}$ it is true that $a_m \leq a_n$ and that there is a number $r \in \mathbb{R}$ such that $a_m \leq r \leq b_n$.
I've proceeding as follows:
We have $a_{n} \leq b_{n} \implies a_{n+1} \leq b_{n+1}$ and thus $a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$.
Does this not imply that $a_m \leq a_n$? Even without stating the obvious fact that the sets are upper and lower bounds of each other? It seems then that $r$ would follow..
EDIT: Taking some of the ideas from below I have written a simple proof. Feedback is welcome and appreciated.
Since $a$ is monotonically increasing and $b$ is monotonically decreasing we have $\forall m,n \in \mathbb{N}, a_n \leq a_{\max(m,n)} \leq b_{\max(m,n)} \leq b_n \implies a_m \leq b_n$. Take $r = a_{\max(m,n)} \text{ or } r = b_{\max(m,n)} \implies a_m \leq r \leq b_n$.

It is clear that $a_n$ is a monotonically increasing sequence bounded above by $b_1$.Hence by Monotone Convergence Theorem $a_n\to r$ (say )
Since you have already proved that $a_n\leq b_n\forall n\in \mathbb N$ it follows that $r\leq b_n\forall n$ 
Hence $a_n\leq r\leq b_n$
