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There's a lot going on here, so I'll start with some overview. It looks like the idea of the proof is to split $1,...,n$ into two pieces, where the first piece is of fixed size $C$, and the second piece has average close enough to $L$. The first piece has fixed size but the second piece can be made so big that it comprises "almost all" of the average.

You usually _end_ with $\epsilon$, but in more complicated proofs, like this one, you sometimes have to bound two (or more) different terms to get the final sum less than ε. In this case, we have two pieces, one of size $C$ and one of size $n-C$, so that usually means we want each piece to be under $\epsilon/2$ so their sum is less than $\epsilon$. This answers (1) and (4).

Now we go after (2). Since $x_n \to L$, $x_n - L \to 0$. So the convergent sequence $x_n - L$ must be bounded by some number $M$. (If you're confused why a convergent sequence must be bounded, you can ask that question separately.) You'll see why we need this later. If you were trying to come up with the proof yourself, you might skip this part for now, and then later realize you don't know anything about the sizes of $x_1$ through $x_{C-1}$. Or perhaps you intuitively realize that in an average, any fixed finite number of terms can be made "insignificant"; to make this precise you would use the bound $M$.

Subtracting the $L$ comes about because we already know two inequalities relating to $|x_j - L|$. It is always less than $M$, and if $j > C$ then it is less than $\epsilon/2$. So if you take the expression for the average, $\left|(x_1 + x_2 + ... + x_n)/n - L\right|$, there are two pieces with that sum: one from $x_1$ to $x_{C-1}$, where each term is within $M$ of $L$, and another from $x_C$ to $x_n$, where each term is within $\epsilon/2$ of $L$. The $nL/n$ is just algebraic manipulation to put a $-L$ on each of the $x_j$ terms.

Finally, you ask about this $N_2$. Notice that $(C-1)M$ is a fixed constant. So if we divide by a "large enough" $n$, then $(C-1)M/n$ can be as small as we want. 

(The comment about $\leq$ vs $\lt$ is mostly pedantic; I think you are right that $<$ would be appropriate in the second part.)