On the monotone class theorem In Real analysis for graduate students by R F Bass,

he proves that the intersection of two sets, either both chosen from $M$ or one is chosen from $M$ and the other $A_0$, belongs to $M$ again (by introducing sets $N_2$ and $N_3$). How proving for two sets implies for a countable intersection to make $M$ a $\sigma$-algebra? I also believe not all elements of $M$ are in a single chain of sets.

I'm not completely sure if this is what you are asking for. But every monotone class $\mathcal{M}$ that is closed under complements and finite intersections must be a $\sigma$-algebra.
You can see it like this: Let $A_1, A_2, \ldots \in \mathcal{M}$. Then we also have
$$B_n := \bigcap_{k=1}^n A_k \in \mathcal{M}$$
for all $n\geq 1$, because $\mathcal{M}$ is closed under finite intersections (*). Furthermore we clearly have $B_1 \supseteq B_2 \supseteq B_3\supseteq \ldots$ so it follows that
$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty B_n \in \mathcal{M}$$
(using that $\mathcal{M}$ is a monotone class).
Hence $\mathcal{M}$ is also closed under countable intersections (and complements, and thus also countable unions), and is therefore a $\sigma$-algebra.

(*) Or, if you only want to use intersections of two sets, do it inductively:
$B_1 = A_1 \in \mathcal{M}$ and $B_n \in \mathcal{M} \implies B_{n+1} = B_n\cap A_{n+1} \in \mathcal{M}$.
