6. For the set $S=\left\{\left(a_{1}, a_{2}, \cdots, a_{5}\right) \mid a_{i}=0\right.$ or $\left.1, i=1,2, \cdots, 5\right\}$, for any two elements $A=\left(a_{1}, a_{2}, \cdots, a_{5}\right)$ and $B=\left(b_{1}, b_{2}, \cdots, b_{5}\right)$, define their distance as: $d(A, B)=\left|a_{1}-b_{1}\right|+\cdots+\left|a_{5}-b_{5}\right|$. Take a subset $T$ of $S$ such that the distance between any two elements in $T$ is greater than 2. How many elements can the subset $T$ contain at most? Prove your conclusion.
6. $T$ contains at most 4 elements.

Assume there is a subset with more than 5 elements that also meets the condition. Then among these 5 elements, at least 3 of them have the same first digit. Let's denote these three elements as $A$, $B$, and $C$, with their first digits being the same.

Similarly, in $A$, $B$, and $C$, for the second, third, fourth, and fifth digits, at least two elements have the same corresponding digit in each position. However, there are only 3 pairs when $A$, $B$, and $C$ are grouped in pairs, so at least two elements, in addition to having the same first digit, also have at least two other digits the same. Let's assume these two elements are $A$ and $B$. Then the distance between $A$ and $B$ is no more than 2, which is a contradiction. Therefore, $T$ has no more than 4 elements.

We can let $T=\{(1,1,1,1,1),(0,0,0,1,1),(1,1,0,0,0),(0,0,1,0,0)\}$. It is not difficult to verify that the conclusion holds. Therefore, $|T|_{\max }=4$.