Prove using induction that from a set of $n+1$ numbers from $\{1,2,..2n\}$, at least one number will evenly divide another.
$3$.
If $n+1$ is also in $A$, then $n+1|2n+2$.
If $n+1$ is not in $A$, then if we replace $2n+2$ with $n+1$, say the new set is $B$, we can use the induction hypothesis like in case $2$ to conclude there exist $a|b$ in $B$. If both $a,b\neq n+1$, then $a,b$ are in $A$ and we are done. If either $a,b=n+1$ then the other is in $A$ and that number divides $2n+2$. $\blacksquare$