Define $S:=\{J\triangleleft R\ : J\subsetneq R\}$ and consider the poset $(S,\subseteq)$. $S\ne\emptyset$ since the trivial ideal $\{0\}\in S$. The conditions for Zorn's lemma are satisfied here because if $C$ is a chain of $S$, the union $\bigcup\limits_{J\in C}J$ is an ideal and an upper bound.
By the lemma $\exists M\in S$ a maximal element. Now show that $M$ is a maximal ideal:
(I'll skip the rest of the proof)
I understand all the process in this proof. But there is one question. For chain $C=\{I_i\}$, $I_1\subseteq I_2\subseteq I_3\subseteq ... \subseteq I_n.$ Then, can I choose an upperbound of each chain as $I_n$?
Almost. Note that there is no guarantee that the chain ever ends, though.
So with a chain which never ends, what can you do? You can take the union of all the ideals in the chain, which is what they do in the proof. You will have to prove that this union is contained in $S$ (i.e. that it is an ideal and that it is proper), and then you can conclude from Zorn's lemma that there are maximal elements.
Note that this only proves that there are maximal ideals. If you want to prove that any ideal $I$ is contained in a maximal ideal, then you would have to restrict $S$ to proper ideals which contain $I$. Exactly the same arguments hold, apart from $I\in S$ being the witness that $S$ is non-empty, rather than $(0)$.