Induction to prove that a set of $n+1$ integers between $1$ and $2n$ has at least one number which divides another number in the set [duplicate]

If for the $n+2$ you select only one of them is $2n +1$ or $2n+2$ then the remaining $n+1$ you picked are from $1... 2n$ and must have one number divide another.

If both of $2n+1$ and $2n+2$ are chosen then assume, for sake of a proof by contradiction, that none of the $n+2$ terms have any term dividing another... then you did not pick $n+1$ and none of the terms divide $n+1$ (which divides $2n + 2$). Discard $2n+1$ and $2n+2$ and pick $n+1$ instead. You now have $n+1$ terms from $1... 2n$. None of the terms that aren't $n+1$ divide $n+1$ or each other. $n+1$ doesn't divide any number less then or equal to $2n$. So none of the terms divide any other. That's a contradiction.

==== old answer where I made the faulty assumption that the terms had to be consecutive ===

Suppose you selected $\{m,m+1,.....m+n+1\}\subset \{1....2n\} $

Then $\{m,.....,m+n\}\subset\{1....2n\} $ has a number that divides another.

If $\{m,m+1,...m+n+1\}\not \subset \{1....2n\} $ then $2n+1 \in \{m,m+1,...m+n+1\}$.

Either $m+n+1=2n+1$ so $m=n$ and the set has $n $ and $2n $,

or $m+n+1=2n+2$ in which case $m=n+1$ and the set has $n+1$ and $2n+2$.