As explained by Greg Martin in a comment:

Certainly $𝐵(𝑥,𝑟)$ is a decreasing function of $𝑟$, but that doesn't imply that the bound on $𝐵(𝑥,𝑟)$ continues to hold for $𝑟\geq 2$.

The reason that $𝑅<2$ is required in the given proof is that it proceeds via upper bounds for $$ \sum_{n \leq x} \Omega(n), $$ and this sum genuinely changes character when $𝑟>2$, since the integers $𝑛$ that are powers of 2 (or nearly so) are then increased by the map $n \mapsto \Omega(n)$ ; for example, the largest power of $2$ less than $𝑥$ already gives a contribution of

$$x^{(\log r)/\log 2}$$ to the sum.

Edit: My only goal is to mark this as answered.

As explained by Greg Martin in a comment:

Certainly $𝐵(𝑥,𝑟)$ is a decreasing function of $𝑟$, but that doesn't imply that the bound on $𝐵(𝑥,𝑟)$ continues to hold for $𝑟\geq 2$.

The reason that $𝑅<2$ is required in the given proof is that it proceeds via upper bounds for $$ \sum_{n \leq x} \Omega(n), $$ and this sum genuinely changes character when $𝑟>2$, since the integers $𝑛$ that are powers of 2 (or nearly so) are then increased by the map $n \mapsto \Omega(n)$ ; for example, the largest power of $2$ less than $𝑥$ already gives a contribution of

$$x^{(\log r)/\log 2}$$ to the sum.