Let $G=\lim_{n\rightarrow\infty}G_n$ be a directed limit of simple groups (e.g. $A_\infty=\lim_{n\rightarrow\infty}\textrm{Alt}(n)$ where the maps in limit are the natural inclusions). Is it true that $G$ is always simple?
No nontrivial homomorphism can arise from one of the $G_n$. There are only two homomorphisms with domain $G_n$, the identity homomorphism, which will lift to the identity on $G$ and the map $G_n\rightarrow\{1\}$, which lifts to $G\rightarrow\{1\}$.
Yes:
Suppose $1\ne H\trianglelefteq G$, so there is some $1\ne x\in H$.
In particular, for some $N$ we have $1\ne x\in G_N$ (identifying each $G_n$ with its image in $G$). For any $n\ge N$ we have that the normal closure of $\langle x\rangle\subseteq G_n$ is $G_n$.
Hence $G_n\le H$ for all $n$, which means $H=G$.