Show that if $T$ is a subset of $S$ having more than $16$ elements then $T$ contains two elements whose distance is at most $2$. 
Let $S = \{0000000, 0000001, ... , 1111111 \}$ be the set of all
  binary sequences of length $7$. The distance of two elements $s_1
,s_2 \in S$ is the number of places in which $s_1$ and $s_2$ differ.
  Show that if $T$ is a subset of $S$ having more than $16$ elements
  then $T$ contains two elements whose distance is at most $2$.

For example, $0001011$ and $1001010$ have distance $2$, since they differ in positions $1$ and $7$.
This problem is not clear to me. I think it's possible to take the pigeonhole principe.
Does someone could help me solve the problem? OR even just a hint would be enough?

Each word has exactly $7$ words at distance $1$, so it "occupies" a total of $8$ words. $17\cdot 8>128$.
