Proof verification for limit problem. If $s_n \leq t_n \leq u_n$ for all $n$, $s_n \rightarrow a \lt b$, and $u_n \rightarrow b,$ then $\lim_{n\rightarrow \infty}t_n \in [a,b].$
Here's how I did it:
Choose $N \in \mathbb{N}$ such that $\forall \epsilon \gt 0$, $n \geq N \rightarrow |s_n-a| \lt \epsilon$ and $|u_n-b| \lt \epsilon$, or
$$-\epsilon+a \lt s_n \lt \epsilon +a$$
$$-\epsilon+b \lt u_n \lt \epsilon +a$$
Hence, $-\epsilon +a \lt t_n \lt \epsilon +b$.
Therefore, $\lim t_{n \rightarrow \infty} \in (a,b)$.
I don't really see how $\lim_{n \rightarrow \infty}$$t_n \in [a,b].$
Doesn't the definition of limit doesn't allow $s_n =a$ and $u_n =b$?
 A: Zeroth, you need to know that the sequence corresponding to $t_n$ actually converges. Otherwise, you might have, for example, $s_n = -1$ and $u_n = 2$ for all $n$. If $t_n$ now alternates between $0$ and $1$, all of your initial constraints are satisfied, but $\lim_{n \rightarrow \infty}t_n$ does not exist.
So: Let us assume henceforth that this limit does exist.
First, to respond to the end of your question: It could be that $s_n$ and $t_n$ correspond to the same sequences, in which case $t_n \rightarrow a$. Similarly, $t_n$ and $u_n$ could correspond to the same sequences, in which case $t_n \rightarrow b$. 
So: You really do need this to be the closed interval $[a,b]$.
Second, you have an error where you write:

Choose $N \in \mathbb{N}$ such that $\forall \epsilon > 0$ ...

You cannot, in general, pick a satisfactory $N$ for all epsilon at once. Rather, for any fixed $\epsilon > 0$ you can pick an $N_1$ to make $s_n$ close to $a$ when $n > N_1$, and similarly pick an $N_2$ for $u_n$ to be close to $b$ when $n > N_2$. Once you have that down, you may wish to pick $N = \max\{N_1, N_2\}$ in order to satisfy both of the inequalities that you then wrote out.
This will, as desired, complete the proof: You have shown that, for any $\epsilon > 0$, you can find an $N$ such that $n > N$ implies $t_n > a - \epsilon$, hence the limit of $t_n$ is $\geq a$; similarly, you have shown that this limit is $\leq b$, which puts the limit in the closed interval $[a,b]$.
