Suppose $(s_n)$ converges and that $s_n \geq a$ for all but finitely many terms, show $\lim s_n \geq a$ I've got a few questions about the problem.

Prob :Suppose $(s_n)$ converges and that $s_n \geq a$ for all but finitely many terms, show $\lim s_n \geq a$

The  solution here breaks this problem up into two parts. 
Q1. I don't understand why is it necessary to consider the finitely many terms that $s_k < a?$
Doesn't the condition that $s_n \geq a$ for all, but finitely many terms and $(s_n)$ being convergent imply that $$\forall \epsilon > 0, \exists N_0 \in \mathbb{N} \implies \forall n > N_0 \implies |s_n - s| <\epsilon$$
So that all the terms after $N_0$ are going to be close to $s$ and therefore $$a < s+\epsilon$$
So why do we need to show the existence of $N > \max \{ M, N_0 \}$ when the definition says there is going to be an $N'$ that gives us convergence?
Q2. Also what is wrong with the following "proof"?

Proof Since $(s_n)$ converges, $$\lim s_n \geq \lim a = a.$$

Ii want to say the proof is wrong because $s_n \geq a$ is not true for every $n$, if it is true for all $n$ then it may be correct? But I thought limits only care about what happens in the long term, so I am not entirely sure what is really the mistake...
EDIT: I was just going to say (very roughly) $s_n \to s \implies \forall \epsilon >0, \exists N_1 \ni \forall n > N_1 s_n < s + \epsilon.$
Now for the rest of the $s_k < 1$ (where $k > N_1$), choose a new $N > \max \{ \max_{k}, N_1 \}$ so that $a \geq s_n < s + \epsilon.$ This is true for all $\epsilon >0$, so take $\epsilon = 0$
 A: Q1. I might be mistaken, but I think this is necessary (at least, for a rigorous proof) because $\forall n > N_0 $ $| s_n - s| < \epsilon$ allows $ 0 < s - s_n  < \epsilon \Leftrightarrow s_n < s$; while we know that there exists such $N_0$ that $s_n$ is close to $s$ when $n > N_0$, nothing assures that this $N_0> i$ (for those finitely many $i$ such that $s_i < a$), so why it would be that $\lim_{n\to\infty} s_n \geq a$? At least, I can't see why this would immediately follow. We have to deal with those $s_i$, and show that they don't matter. For one possibly way of doing that which differs a little from the solution you linked to, see for example marty's answer).
EDIT. Okay, I think that works, too (quite elegantly, even). However, I'd argue that it is a substantial step in the proof: the main difference (and point) between this theorem and the more basic one mentioned below ("suppose $s_n > a$ for all $n$", which is usually proved in the textbooks) is that for finite number of elements $s_n < a$, and showing why yet $\lim_{n\to\infty} s_n > a$. In the solution you linked this is dealt with a proof by contradiction, but you already noted the similarities in the argument in the comments. 
Someone more proficient than me might be able to provide a more refined explanation why to address this step with a care (or disagree with me), but to me it strikes as a natural thing to do (in other words, the point of proving this theorem as a separate exercise in the first place). (end of edit.)
Q2. Well, I believe you're correct in both of your sub-questions: You can't deduce (in this straightforward fashion) that $\lim_{n\to\infty} s_n \geq \lim_{n\to\infty} a = a$, if for some $i$, $s_i <a$, but if $s_n > a $ for every $n$, then it is straightforward to show that $\lim s_n > \lim a = a$.
Generally speaking, you've got a right idea in that "limits only care about what happens in the long term", but (1) as a statement, it's quite a handwave-y one and (2) that result has to be established before you can use it, which brings us back to Q1.
A: Suppose $\lim s_n < a$.
Let $L = \lim s_n$.
Since $L < a$,
if $c = a-L$,
then $c > 0$.
Since $L = \lim s_n$,
by the standard definition of limit,
there is an $N$ such that
$|L-s_n| < c/2$
for all $n > N$.
Rewrite this as
if $n > N$,
$-c/2 < L-s_n < c/2$
or
$s_n < L+c/2
=L+(a-L)/2
=(a+L)/2
< a
$
since $L < a$.
Therefore
$s_n < a$
for all but a finite number of terms.
Restating,
if $\lim s_n < a$,
the
$s_n < a$
for all but a finite number of terms.
By good old contrapositive,
if
$s_n \ge a$
for all but a finite number of terms
then
$\lim s_n \ge a$.
