Worth mention: there are various (folklore) forms of Zorn's Lemma that prove more convenient for these type of inductive arguments. Below is one such, excerpted from: $ $ Anderson; Dobbs; and Zafrullah: Some applications of Zorn's lemma in algebra.
The use of Zorn’s Lemma has been a part of the mainstream in virtually every
area of algebra for more than $75$ years. The aim of this note is to indicate some new
applications of Zorn’s Lemma to a number of algebraic areas by using a slightly
different perspective. In Theorem $1$ and its aftermath, we show that a property $P$
holds for all the subobjects of a given object if and only if $P$ supports both the chain
condition from Zorn’s Lemma and some finitistic conditions on subobjects that have
the flavor of mathematical induction.
Theorem $1$. $ $ Let $R$ be a ring and let $E$ be a (left) $R$-module. Let $P$ be a property
of modules. Then each nonzero submodule of $E$ satisfies property $P$ if and only if
the following $3$ conditions hold.
$(1)\ \ P$ is true for every nonzero cyclic submodule of $E$.
$(2)\ \ $ If $A$ is a nonzero submodule of $E$ such that $A$ satisfies property $P$ and $x\in E$, then $A + Rx$ again satisfies property $P$.
$(3)\ \ $ If $\{H_\alpha\}$ is a chain of nonzero submodules of $E$ such that each $H_\alpha$ satisfies
property $P$, then $\cup H_\alpha$ satisfies property $P$.
Proof. $ $ Suppose that $(1)$, $(2)$ and $(3)$ hold. If the “if” assertion fails, $E$ has a nonzero
submodule $A$ such that $A$ does not have property $P$. Let $S$ be the set of all nonzero
submodules $I$ of $A$ such that $I$ has property $P$. Then $S$ is nonempty because, by
$(1)$, every nonzero cyclic submodule has property $P$. Moreover, $S$ can be partially
ordered by inclusion. By $(3)$, S meets the requirements of Zorn’s Lemma, and so
must contain a maximal element $J$. Since $A$ does not have property $P,\ J \subsetneq A$.
Pick $\,x \in A\backslash J$. Note that $J + Rx \subseteq A$ has property $P$ (by $(2)$), contradicting the
maximality of $J$, thus completing the proof of the “if” assertion. The converse is
evident. $\ \small\rm QED$
Remark $2$. $ $ In the statement of Theorem $1$, we can replace $(2)$ by
$(2')$: $ $ If $A$ and $B$ are two nonzero submodules of $E$ such that $A$ and $B$ satisfy property $P$, then
$A+B$ satisfies property $P$. We will emphasize the case where $P$ holds for all nonzero
submodules and leave the case where it holds for all nonzero proper submodules to
Theorem 3 and the comment that precedes it.
Recall that a (left) $R$-module $M$ is semisimple if $M$ is a direct sum of simple
submodules (see, e.g., [5, page 11]). A semisimple module $M$ is characterized by
the property that every submodule of $M$ is a direct summand of $M$ [5, Proposition 4.1, page 11]. Of course this characterization does not change if we replace
“every submodule” by “every nonzero submodule”. Also, as noted in the proof of
[5, Proposition 4.1, page 11] (cf. also [4, Theorem 2.3.2]), every submodule of a
semisimple module is a semisimple module. Setting $P$ = “is a direct summand of
$M$” in Theorem $1$ gets us the following new characterization of semisimple modules.
Proposition $5$. $ $ An $R$-module $M$ is a semisimple $R$-module if and only if the
following $3$ conditions hold.
$(1)\ \ $ Every nonzero cyclic submodule of $M$ is a direct summand of $M$.
$(2)\ \ $ If $A$ is a nonzero submodule that is a direct summand of $M$, then for every
nonzero cyclic submodule $B,\ A + B$ is a direct summand of $M$.
$(3)\ \ $ If $\{H_\alpha\}$ is a chain of nonzero submodules of $E$ such that each $H_\alpha$ is a direct
summand of $M$, then $\cup H_\alpha$ is a direct summand of $M$.
Proof. $ $ That $M$ semisimple implies $(1),\, (2)$ and $(3)$ is clear from the above comments. For the converse, take $P$ = “is a direct summand of $M$” in Theorem $1$. Combining this with the fact that the zero submodule is trivially a direct summand of $M$, we have the conclusion that $M$ is semisimple. $\ \small\rm QED$