At page 460 of the book Algebra by Dummit it is written that $N \subseteq L$ implies $\mathrm{Ann}(L) \subseteq \mathrm{Ann}(N)$.
Recall that for any submodule $N$ of $M$, the annihilator of $N$ is the ideal of $R$ defined by $\mathrm{Ann}(N) = {\{r \in R \ | \ rn = 0 \ \ \ \text{for all} \ n \in N}\}$.
Suppose $a\in L-N$. Therefore there is some $r \in R$ such that $ra=0$. This $r$ doesn't need to be in $\mathrm{Ann}(N)$ then $\mathrm{Ann}(N) \subseteq \mathrm{Ann}(L)$ but not the reverse!
What is wrong with my argument and how $N \subseteq L$ implies $\mathrm{Ann}(L) \subseteq \mathrm{Ann}(N)$?
The implication is simple enough. Because $N\subseteq L$, the condition "$nr=0$ for all $n\in N$" is less restrictive than "$nl=0$ for all $l\in L$": There are fewer elements that $r$ has to kill in order to be part of the annihilator of $N$. Therefore it is possible that there are more $r$ that fulfills that condition. At the very least, an $r$ that kills every $l\in L$ must kill any and all $n\in N$.
The problem with your approach is that the set that you argue is smaller than $\operatorname{Ann}(N)$ is really $\operatorname{Ann}(a)$. You're only checking whether the relevant $r$ kills $a$ (note that those two annihilators are generally incomparable, since there might just as well be $r$ such that $ra\neq0$ and $rn=0$ for all $n\in N$).
But for $r$ to be an element of $\operatorname{Ann}(L)$, it has to kill every element of $L$, including $a$, but also including the rest of $L-N$ as well as all of $N$.