Well I have used spheres in coding with radius, $r=\left\lfloor\frac{\delta -1}{2} \right\rfloor=\left\lfloor\frac{8 -1}{2} \right\rfloor = 3$
and that means we have $\sum \limits_{i=0}^3 {16 \choose i} (2-1)^i=697$ words in each sphere, with $2^k=2^5=32$ spheres we have $32*697=22304$ distinct words.
Since we can have a total of $2^{16}=65535$ such a code does exist
Is this an acceptable proof?
Such a code exists. Others have posted links and/or construction schemes. I would just state that the Reed-Muller code $RM(1,4)$ is exactly what you need.
My main point is to explain, amplifying the message from Dan Uznanski and Pieter21, why your argument is insufficient (it could lead to a non-existence proof though). Assume that the question were about the existence of a $(7,1,8)$-code. Following your logic we would calculate $$ r=\left\lfloor\frac{\delta-1}2\right\rfloor=3, $$ and that the number of words in a ball of radius $r$ is $$ V(B(r=3))=\sum_{i=0}^3\binom 7 i=1+7+21+35=64. $$ You would then conclude that the space $GF(2)^7$ has room for $2=2^k$ balls.
But the minimum distance of a code obviously cannot exceed the length, so a $(7,1,8)$-code clearly cannot exist.
It was a bit naughty of me to do it that way, sorry. Also there exists a $(7,1,7)$-code, so my example for showing the error of your ways is not as convincing as it might be given that it depended on $\delta=8$ and $\delta=7$ producing the same $r$. The main reason why the point count is not sufficient is that you would still need to prove that you can actually place the balls in such a way that there is no overlap, just as Dan said.