I just want to confirm my proof is error free, since it seems quite different from others I've come across.
Consider $\{\sigma,...,2n\}$ for $\sigma \in \{1,...,n\}$. Obviously we have:
$n \mid 2n$ , $n-1 \mid 2n-2$ , . . . , $\sigma \mid 2\sigma$
For all $x,y\in \{\sigma,...,n\}$ we know $2x \ne 2y$ unless $x=y$. Thus if $2x \notin \{\sigma,...,n\}$ then at least one novel member of every such pair $\{x,2x\}$ must be deleted from $\{\sigma,...,2n\}$ to prevent having an entry and its double in $\{\sigma,...,2n\}$. If instead $2x\in \{\sigma,...,n\}$ then we must delete at least two members (at least one novel) of every such triple $\{x,2x,4x\}$ from $\{\sigma,...,2n\}$ to prevent having an entry and its double and / or quadruple in $\{\sigma,...,2n\}$.
This means we must remove at least one novel element $\textit{for}$ each member of $\{\sigma,...,n\}$, which amounts to removing at least $n-\sigma+1$ elements from $\{\sigma,...,2n\}$ thus leaving us with less than $n+1$ elements if we are to have no elements which divide each other.
This proof follows a generally bad pattern that can be described as "Let me think of the worst thing that could happen, and then show that it doesn't happen." Such proofs have two perils:
You may think of a bad thing, but fail to think of the worst, and because the proof doesn't contain a proof that your situation is the worst, no one (especially you) ever notices it.
The second-worst thing might also provide a counterexample if you're not careful.
If you want to show that for every subset $S$ of size $n_+1$, some property holds, your proof should almost certainly start with "Let $S = \{x_0, \ldots, x_n\}$ be a subset, where the $x_i$ are all distinct, so that $S$ has $n+1$ elements. We now show that ..."
Saying "Let $S$ be the subset containing the top $n+1$ elements" makes the assumption that this is the "worst case", and you're headed down a very bad path.