I came across the following exercise in my algebra textbook.
Let $M$ be a left $R$-module ($R$ is ring with unity $1_R$) and let $N\subseteq M$ be a submodule such that $x\in M-N$ for some $x\in M$. Show $M$ has a maximal submodule $L$ such that $L\supseteq N$ and $x\not\in L$.
I guess I must use Zorn's lema.
Sketch: I defined $$\mathscr{F}:=\{L\subseteq M: L\neq M\ \textrm{is submodule}; L\supseteq N\quad \textrm{and}\quad x\not\in L\}.$$ Then $\mathscr{F}\neq \phi$ for $N\in\mathscr{F}$. Furthermore $(\mathscr{F}, \subseteq)$ is a poset. Given a totally ordered subset $\mathscr{T}$ of $\mathscr{F}$, it is easy to see $$L_0:=\bigcup_{L\in\mathscr{T}} L,$$ is an upper bound for $\mathscr{T}$. Hence $\mathscr{F}$ has a maximal element $\overline{L}$. In particular, $\overline{L}$ is a submodule of $M$, $\overline{L}\neq M$, $\overline{L}\supseteq N$ and $x\not\in \overline{L}$.
I'm almost there, how to show $\overline{L}$ is maximal with respect to the desired property?
This is not true for arbitrary modules over unital rings. An example is $\mathbb Q$ as a module over $\mathbb Z$, which does not have a maximal submodule (see There are no maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$).
In your proof you have not shown that $L_0\neq M$, which is the source of this contradiction.