Order of Elements in Quotient Groups (Proof) Let G be a group and N $\lhd$ of G such that [G : N] = 100. Suppose that a $\in$ G such that $a^{23}$ = e, where e is the identity element of G. Show that a $\in$ N.
Since the order of the group G/N is 100 then $(aN)^{100}$ = N since $a^{100}N$ = $N$ which implies that $a^{100}N \in N$
Since $a^{23}$ = e and $a^{23} \subseteq a^{100}$, ($\Rightarrow$) $a^{23}N = N$ ($\Rightarrow$) $eN = N$. Thus, $a \in N$
Does this proof make sense or am I missing something?

There are some issues with the argument. You can't say $a^{100}N \in N$ since $a^{100}N$ is a set but you can say $a^{100} \in N$, which is the correct implication of $a^{100}N = N$. It does not make sense to say $a^{23} \subseteq a^{100}$ since $a^{23}$ and $a^{100}$ are elements of $G$, not sets. 
Since you know that $a^{23} = e$, you have $(aN)^{23} = a^{23}N = eN = N$. Thus, the order of $aN$ divides $23$. Since $G/N$ has order $100$, the order of $aN$ also divides $100$. Thus $|aN|$ divides $gcd(23, 100)$, which equals $1$. Thus, $aN = N$, which shows that $a \in N$.
