Let the initial state be $A_0 = (bB, wW)$, and let $A_{i-1} \to A_i$ denote performing the $i$-th "arrange-split" operation.
Following the operation pattern, until all piles consist of a single ball. We mark the following important stages:
$A_s$ : the first time a pile of a single ball appears (whether black or white).
$A_t$ : when the total number of piles exceeds $K$; however it is possible that: when all piles are single balls the total is still no more than $K$ piles, in which case let $t = \infty$.
$A_f$ : when all black balls are single-ball piles.
We only need to prove: the stopping moment required by the problem occurs at $A_{f-1}$ or earlier, that is, in $A_{f-1}$ there must be at least one pile of a single white ball.
Clearly, $s \le f$. (*) Also in $A_{f-1}$ the black balls must have some piles that are $b2$ (ranked among the first $K$), and all others are $b1$.

When $i < \min\{t, s\}$, every pile in $A_i$ is selected, and must be split into two (since there is no single-ball pile yet). Suppose at this point the maximum and minimum numbers of the black piles are $M_i$ and $m_i$ respectively, and the maximum and minimum numbers of the white piles are $N_i$ and $n_i$ respectively. For $A_i$, we have the following properties:
(1) The total number of black piles and the total number of white piles are both $2^i$.
(2) $M_i \ge N_i$.
(3) $m_i \ge n_i$.
All of these can be easily proved by induction.

Regarding the relative sizes of $s, t, f$, there are two possible cases:

Case 1. [$s \le t$ or $f \le t+1$] In particular $t = \infty$ belongs to this case. Suppose $s=f$ occurs. By the definition of $s$, $m_i, n_i \ge 2$. By (*) we get $M_{s-1} \le 2$. So every pile has exactly two balls. But this contradicts $B > W$ together with property (1)! So it must be that $s \le f - 1$.
The given condition [$s \le t$ or $f \le t+1$] combines into: $s \le f - 1 \le t$. Consider $A_{s-1} \to A_s$. Since every pile in $A_{s-1}$ is selected and $m_{s-1} \ge n_{s-1}$, when splitting into $A_s$ we have $m_{s-1} \to (m_{s-1} - 1, 1)$, i.e., $m_s = 1$; simultaneously $n_{s-1} \to (n_{s-1} - 1, 1)$, i.e., $n_s = 1$; that is, the single-ball black pile and the single-ball white pile first appear simultaneously at $A_s$. Since $s \le f - 1$, the stopping moment required by the problem occurs at $A_{f-1}$ or earlier.

Case 2. [$t+1 \le s$ and $t+2 \le f$] At $A_{t-1}$ the total number of piles is $2^t$, so $2^t \le K < 2^{t+1}$, and at $A_t$ the total number of piles is $2^{t+1}$ (black and white each being $2^t$). Perform step $t+1$: $A_t \to A_{t+1}$, among the selected $K$ piles at most $2^t$ can be black piles, so $A_{t+1}$ has at least $2^t + (K - 2^t) = K$ white piles. Since the total number of white piles is non-decreasing, the number of white piles in $A_{f-1}$ is at least $K$.

Finally, in $A_{f-1}$, $M_{f-1} = 2$, and these $b2$ piles (at least one) are all split into $(1, 1)$ at step $f$, so in $A_{f-1}$ at most $K-1$ white piles can be selected, meaning there is at least one white pile $\Omega$ that is not selected. This pile $\Omega$ has exactly one white ball, because if pile $\Omega$ had more than one white ball, then $\Omega$ should be ranked before the $b2$ piles, and thus it should have been selected, which is a contradiction! Therefore, in $A_{f-1}$ there must be at least one pile with a single white ball, which completes the proof.