How many elements of order 7 in a simple group of order 168? I'm not sure my proof is correct because I don't use the fact that the group is simple anywhere...

Since $168=2^3 \cdot 3 \cdot 7$, there exists a subgroup of order 7. Let $n_7$ be the number of Sylow 7-subgroups. Since $n_7 | 24$ and $n_7 \equiv 1$ mod $7$, $n_7 = 8$. Since every group of prime order is cyclic, there exist 8 elements of order 7.  

I think that I still need to prove that there are no other elements of order 7 and that's where the fact that the group is simple would come in.
The $8$ Sylow subgroups of order $7$ each contain $6$ elements of order $7$ together with the identity. Since an element of order $7$ generates a group of order $7$ these elements are all distinct. There are $8 \times 6=48$ elements of order $7$.

You use the fact that the group is simple in asserting that there are $8$ subgroups of order $7$. It is always possible to have $1$ subgroup of a given order (so long as it is a factor of the group order e.g. the cyclic group of order $168$) - but a single subgroup of order $7$ would be normal, and the group would not be simple.
168=8⋅3⋅7 
the elements of sylow 7-groups of grop of order 168 
the order of 7 subgroups is 7. number of elements between 0-168 which is 1 mod 7 are 2,8,15,22,29... 
since 1 and 8 are the only positive integers congruent to 1 mod 7 and divides 168, the number of sylow 7-subgroups is 8. 
then there are 8x6=48 elements of order 7 since identity is of order 1 and common in all the 8 sylow 7-subgroups