Let $(s_n)$ be a sequence that converges... Exercise 8.9 from Elementary Analysis: The Theory of Calculus by Kenneth A. Ross:  
Let $(s_n)$ be a sequence that converges.  
(a) Show that if $s_n \geq a$ for all but finitely many $n$, then $\lim s_n \geq a$.
(b) Show that if $s_n \leq b$ for all but finitely many $n$, then $\lim s_n \leq b$.
(c) Conclude that if all but finitely many $s_n$ belong to $[a, b]$, then $\lim s_n$ belongs to $[a, b]$.

Here's what I have so far:
Consider the set $S = \{n \in \mathbb N : s_n < a\}$. By assumption, $S$ is a finite nonempty subset of $\mathbb R$. So it must have a maximum, say $M$. Now assume $s = \lim s_n$ and $s < a$. Then for each $\epsilon > 0$, there exists $N_0 \in \mathbb N$ such that $$n > N_0\; \text{implies} \mid s_n - s\mid < \epsilon\text{,}$$ or, in the case where $\epsilon = a - s > 0$,   $$n > N_0\; \text{implies} \mid s_n - s\mid < a - s \text{.}$$ Define $N := \max \{N_0, M\}$. Then $$n > N\; \text{implies}\; s_n \geq a\; \text{and} \mid s_n - s\mid < a - s \text{.}$$ The latter inequality implies, in particular, $s_n < a$ for all $n > N$. But this is contradiction, so our assumption that $s < a$ must be false. Thus $s \geq a$. $\square$

I was hoping somebody could help me make sense of this. The book suggested drawing a picture, but I couldn't figure out how. So what would this look like on both the  number line and on the $n$-$s_n$ plane? Also, is there a better way to prove part (a)?  

I think making sense of it would involve drawing a kind of Cauchy sequence, like this typical one:

Pretty much what's going on is that, if only finitely many $s_n<a$, then when $n$ is large enough, $s_n\geq a$. This, you may recognize, is the definition of a limit (more or less). In this picture, imagine drawing a horizontal line in the $y-$axis lower than the eventual limit as drawn. At some later $n$, every point in the sequence has to be above the line, which again means the limit is larger than the value.
