show that $S_7$ has no subgroup of order 11 how to show that $S_7$ has no sub group of order 11
Is it a proof by contradiction, whereby, let H be the subgroup of order 11,
assume H is a subset of subset of $S_7$, then by lagrange's theorem we state it is not a subset since [...]

You've got the right idea, but need to flesh it out and arrive at a contradiction.
Assume there exists an $\,H \leq S_7\,$ with $\,|H| = 11$. 
Then by Lagrange's theorem (the order of a subgroup of a finite group divides the order of the group), we have that $\,11\,$ divides $\,|S_7| = 7! = 7\cdot 6\cdot 5\cdot 4\cdot 3 \cdot 2 \cdot 1.$ 
But as we see, that's impossible: $\;11\,{\large \not\mid}\,7!$. 
Therefore, no such subgroup $H$ exists.
