Proof verification: A $(16,5,8)$ binary code does exist. 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?
 A: Your assumption is wrong, at least in the general case. Dan Uznanski already showed it intuitively.
If you want to prove such code exists, you have to construct one.
You can construct a linear binary code of $[16,5,8]$ using following code constructs:
Construction of a linear code 
[16,5,8] over GF(2):
[1]:  [8, 1, 8] Cyclic Linear Code over GF(2)
     RepetitionCode of length 8
[2]:  [4, 1, 4] Cyclic Linear Code over GF(2)
     RepetitionCode of length 4
[3]:  [4, 3, 2] Cyclic Linear Code over GF(2)
     Dual of the RepetitionCode of length 4
[4]:  [8, 4, 4] Quasicyclic of degree 2 Linear Code over GF(2)
     PlotkinSum of [3] and [2]
[5]:  [16, 5, 8] Linear Code over GF(2)
     PlotkinSum of [4] and [1]

source: http://www.codetables.de/
A: Such a code exists. There even exists an optimal binary linear $[16,5,8]$-code. For its construction, see for example exercise $121$ in the book "Fundamentals of Error-Correcting Codes", wriiten by W. Cary Huffman  and Vera Pless. Just to say that we can have $2^{16}$ words does not prove the existence of such a code.
A: 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. 
