Two binary codewords of weight $4$ at distance $6$ from each other must
have a common $1$ in exactly one position, right? So for starters, here
are three such codewords
$$1111000000\\1000111000\\1000000111$$
Can you find two more? You cannot. So let us back up a little bit
and think some more since three codewords sharing a common $1$ in the
same position does not work. What if we chose the five putative codewords
so that for any given codeword, the other four had a common $1$ in
four different positions? So, now we have
$$1111000000\\1000111000$$
as before, but now codeword #3 shares a $1$ in position $2$ with codeword #1 and a $1$ in position $5$ with codeword #2, so that the first three codewords are
$$1111000000\\1000111000\\0100100110$$
The fourth codeword shares a $1$ in positions $3,6,8$ with codewords #1, #2, #3
respectively giving
$$1111000000\\1000111000\\0100100110\\0010010101$$
Can you now figure out the fifth codeword? Which positions have not been used
thus far for sharing?
Note that there are $\binom{5}{2} = 10$ pairs of codewords and each pair
shares a $1$ in a different position.
Thanks to Dilip in the comments, I was able to come up with a solution. I figured I would post a solution for anyone else interested in the problem.
As mentioned above, consider the first-order binary Reed-Muller code, $\mathcal{R}(1, 4)$. This is a $[2^4, 4+1, 2^{4-1}] = [16, 5, 8]$ code. Then, shorten this code by deleting the over-all parity check bit and taking only the remaining codewords of even weight. By doing this, we we are then left with a $[15, 4, 8]$ (which can be verified by looking at the generator matrix for $\mathcal{R}(1, 4)$).
Now, further shorten this code by deleting a nonzero coordinate. We will be left with a $[14, 4, 7]$ code. By shortening the $[14, 4, 7]$ code by deleting another nonzero coordinate, we will get a $[13, 4, 6]$ code.
Best Answer
Since the discussion in the comments between @GitGud and the OP seems to converging far too slowly to a solution, and since I wrote the answer wherein the statement in question occurred, here goes with a few hints for a different approach which will, along the way force you to learn some useful facts about binary vectors.
Suppose $\mathcal C$ denotes a linear binary code. Partition $\mathcal C$ into two subsets $\mathcal C_0$ and $\mathcal C_1$ consisting respectively of all the codewords of even Hamming weight and all the codewords of odd Hamming weight.
Show that for every linear binary code, $\mathcal C_0$ is a non-empty set. Hint: find a codeword of even weight in $\mathcal C$. (Subhint: $0$ is an even integer).
Explain why if $\mathcal C_1$ is an empty set, then we have proved part of the statement in question.
Digression:
Show that the sum of two binary vectors of even Hamming weight is a vector of even Hamming weight.
Show that the sum of two binary vectors of odd Hamming weight is also a vector of even Hamming weight.
Show that the sum of a binary vector of odd Hamming weight and a binary vector of even Hamming weight is a vector of odd Hamming weight.
End of digression
Suppose that $\mathcal C_1$ is non-empty and let $x$ denote a fixed element of $\mathcal C_1$.
Show that $x + \mathcal C_0 = \{z \colon z = x + y, y \in \mathcal C_0\}$ is a collection of $|\mathcal C_0|$ distinct vectors, all of which have odd Hamming weight. Argue that $x + \mathcal C_0 \subset \mathcal C_1$ and so it must be that $|\mathcal C_1| \geq |\mathcal C_0|$.
Show that $x + \mathcal C_1 = \{u \colon u = x + v, v \in \mathcal C_1\}$ is a collection of $|\mathcal C_1|$ distinct vectors all of which have even Hamming weight. Argue that $x + \mathcal C_1 \subset \mathcal C_0$ and hence $|\mathcal C_1| \leq |\mathcal C_0|$.
Conclude, if you dare, that either $\mathcal C_1 = \emptyset$ and so all the codewords in $\mathcal C$ have even Hamming weight, or that $\mathcal C$ contains codewords of even Hamming weight as well as odd Hamming weight and that $$|\mathcal C_1| = |\mathcal C_0| = \frac 12 |\mathcal C|.$$