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 $\epsilon = 0$

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
$\lim s_n \ge a$.
