Let $a$ non-unit. Consider the ideal $\langle a \rangle$. Since $a$ is not a unit this ideal does not contain $1$ so it is not the full ring $R$. Now, let's look at the ideals of $R$ that are not equal to $R$. In particular, $\langle a \rangle$ is one of them. The question is whether $\langle a \rangle $ can be enlarged to an ideal again not $R$ which itself cannot be enlarged any further. OK, if $\langle a \rangle$ itself cannot be enlarged any further, done, it is maximal. If not, then it can be enlarged to say some ideal $I_1$. If $I_1$ is maximal, done. If not, enlarge it to $I_2$. If $I_2$ is maximal, done.
If not, enlarge it to $I_3$, and so on.
What can happen:
one obtains at some point an ideal $I_n$ that is maximal. Done
one never stops, and we get
$$\langle a \rangle \subset I_1 \subset I_2 \subset I_3 \subset \ldots $$
strict inclusions.
Now one can argue that in a lot of commonly met rings this can never happen. This is true, it can never happen in a quite large family of rings, called noetherian. But still, in principle one can go on for ever like this. What to do? Let's look at the union of all the $I_n$. It is an ideal itself, and moreover, it is not $R$. Why? I mean in principle it may just be possible that they exhaust $R$. But, attention. None of the $I_n$ contain the element $1$. This is the key. It follows that the union also does not contain $1$ and hence we got an even bigger ideal, still not $R$. If this is maximal, now we are done. What if not? Well, we should just keep on going and increasing at each step to a large one if we haven't yet reached a maximal. OK, one should be careful: how many steps can we perform? Already we have performed something like countably infinitely many. Well, is there a possibility to keep on going for ever? Certainly one cannot just go on forever in this extended sense, when the number of steps can be labelled by any $\it{well\ ordered \ set}$. Well, this is the content of a tricky method called the Zorn lemma. I will end now by saying that at some (perhaps very far away) point the procedure will stop and we will get our maximal ideal containing $\langle a \rangle$