Prove that if there exists a chain of subgroups $G_1 \leq G_2 \leq \cdots \leq G$ such that $G = \bigcup_{i=1}^\infty G_i$ and each $G_i$ is simple then $G$ is simple.
I have proved the following lemma:
Let $N \triangleleft G$. Then $N\cap G_i \triangleleft G_i$ for each $G_i$.
$N \cap (\cup_{i=1}^\infty G_i) = \cup_{i=1}^\infty (N \cap G_i)$
Since each $G_i$ is simple, we have $N \cap G_i \in \{ 1, G_i \}$.
Suppose $N \cap G_i = 1$ for all $G_i$. Then $$N = N \cap G = N \cap (\bigcup_{i=1}^\infty G_i) = \bigcup_{i=1}^\infty (N \cap G_i) = \bigcup_{i=1}^\infty 1 = 1.$$ Hence $N = 1$.
Then I am not quite sure how to handle the case that $N \cap G_k = G_k$?