It is not hard to prove the following lemma:
LEMMA: Let $(G_i)_{i \in I}$ be a chain of simple groups. Then $G = \bigcup_{i \in I} G_i$ is a simple group.
Let $N$ be a normal subgroup of $G$ and define $N_i = N \cap G_i$ (all $i \in I$). $(N_i)_{i \in I}$ is a chain such that $N = \bigcup_{i \in I} N_i$. Let $n \in N \trianglelefteq G$, then $n \in G_i$, some $i \in I$; then $n \in N \cap G_i = N_i$. Observe that $N_i \trianglelefteq G_i$, a simple group (all $i \in I$). Thus, $N_i = 1$ or $G_i$ (all $i \in I$). Now we have two cases:
- $N_i = 1$ (all $i \in I$). Then $N = \bigcup N_i = 1$
- $N_i = G_i > 1$, some $i \in I$. If $1 < N_i \leq N_j$, we have $N_j \neq 1$ and so $N_j = G_j$ (all $j \in J$). It follows that $N = \bigcup N_i = G$.
QED.
Zorn's lemma now implies the existence of "maximal simple groups." These would be simple groups which cannot be imbedded in any other simple groups. Can we explicitly construct such a group? Or would such a group be purely nonconstructive (if we can even prove such a thing)?
You cannot invoke the Zorn-Kuratowski-Lemma here: The collection of all simple groups is not a partially ordered set; it is a proper class. Therefore, at least your argument for why there must exist a maximal simple group does not work (although such a group might exist by different reasoning, which I do not know).
If Zorn's Lemma would work for partially ordered classes as well, then you would also be able to prove the existence of a maximal group – or even a maximal set – which clearly does not exist. To illustrate the failure, one proof of Zorn's Lemma uses a well-ordering of your poset; this of course requires that you have a set and not a proper class.