Suppose $G$ is a locally finite group such that $G=\bigcup_{i=1}^\infty S_i$, where $S_i$ is a finite group and $S_i \triangleleft S_{i+1}$ for all $i \in \mathbb{N}$. Let $P$ be a Sylow (maximal with respect to inclusion) $p$subgroup of the group $G$. For each $i \in \mathbb{N}$ put $P_i=S_i \cap P$. Then $P_i$ is a Sylow $p$subgroup of $S_i$ for all $i \in \mathbb{N}$. Or not?
Yes. Fix $i$. For $n\ge i$, let $L_n$ be a $p$Sylow of $S_n$ containing $P_n$. Note that $S_i$ is subnormal in $S_n$. By the lemma below, $L_n$, the intersection $M_n=L_n\cap S_i$ is a $p$Sylow of $S_i$. Now there exists an infinite set $N$ of $n$ such that the Sylow $M=M_n$ does not depend on $n\in N$. Since $\langle P_n,M\rangle$ is a $p$group for all $n$ and $P=\bigcup P_n$, we deduce that $\langle P,M\rangle=\bigcup\langle P_n,M\rangle$ (increasing union) is a $p$group. By maximality, we deduce that $M\subset P$, QED. [Lemma: if $G$ is a finite group, $H$ a subnormal subgroup and $P$ a $p$Sylow of $G$ then $P\cap H$ is a $p$Sylow of $H$. Indeed, by an obvious induction we can reduce to the case when $N$ is normal, let $Q$ be a $p$Sylow of $H$. Then $Q$ is contained in a $p$Sylow of $G$, that is, a conjugate $gPg^{1}$. So $g^{1}Qg\subset P\cap H$ (because $H$ is normal); by cardinality this is also a $p$Sylow of $H$ and the lemma is proved.] 

