I am considering the following set $$ C = \{ (0,0,0,0,0,0,0), (0,0,0,1,0,1,1), (0,0,1,0,1,1,1), (0,0,1,1,1,0,0), (0,1,0,0,1,0,1), (0,1,0,1,1,1,0), (0,1,1,0,0,1,0), (0,1,1,1,0,0,1), (1,0,0,0,1,1,0), (1,0,0,1,1,0,1), (1,0,1,0,0,0,1), (1,0,1,1,0,1,0), (1,1,0,0,0,1,1), (1,1,0,1,0,0,0), (1,1,1,0,1,0,0), (1,1,1,1,1,1,1)\}. $$
Is there an easy way to prove that $\bigcup_{x\in C} \{y\in \{0,1\}^7\mid d(x,y)\leq 1\} = \{0,1\}^7$, where $d$ is the Hamming-distance.
Obviously the inclusion "$\subset$" is trivial.
Regarding the inclusion "$\supset$". We choose an arbitrary $z \in \{0,1\}^7$. If $z \in C$, we are done. So suppose $z \notin C$. If $z$ is different in only one bit from any word in $C$, we have $z \in C$ and are done. It remains to show that there is no $z \in \{0,1\}^7$ that differs in two (or more places) from all words in $C$. It there an easy argument to conclude this (only assuming basic knowledge about $d$)?
Any help would be greatly appreciated.
It is fastidious but not difficult to see that the minimal distance of your code is $3$, hence the balls of radius $1$ centered at each codeword are pairwise disjoint. Each ball has cardinality $8 = 2^3$ and there are $16 = 2^4$ of them, for a total of $2^7$ words.