= Solution
Let the $r$ distinct <codewords> be $x^{(1)},\ldots,x^{(r)}\in\mathbb F_2^N$. No <linearity> assumption is needed. Sum their <Hamming distances> over unordered distinct pairs:
$$
S=\sum_{1\leq i<j\leq r}d(x^{(i)},x^{(j)}).
$$
Each summand is at least the <minimum Hamming distance> $\delta$, so $S\geq\delta r(r-1)/2$.
For coordinate $a$, let $m_a$ be the number of <codewords> whose $a$th bit is $1$. Exactly $m_a(r-m_a)$ unordered pairs differ at that coordinate. Summing coordinate contributions gives
$$
S=\sum_{a=1}^Nm_a(r-m_a)
\leq\sum_{a=1}^N\frac{r^2}{4}
=\frac{Nr^2}{4},
$$
since $m_a(r-m_a)=r^2/4-(m_a-r/2)^2$. Combining the two bounds and dividing by $r>0$ gives $2\delta(r-1)\leq Nr$, hence $r(2\delta-N)\leq2\delta$. Under $2\delta>N$, division is legitimate, proving the <Plotkin bound>
$$
\boxed{r\leq\frac{2\delta}{2\delta-N}}.
$$
Because $r$ is an integer, it also satisfies the floor of the right-hand side. The binary alphabet is essential to the coordinate count; this proof applies to arbitrary <binary block codes>, not just linear ones.
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