Block-cluster leaves partition the index-pair set

If a block node $(\tau ,\sigma )$ is split on the row side, its child rectangles $({\tau }_i,\sigma )$ are disjoint and cover $\tau \times \sigma$ because the ${\tau }_i$ partition $\tau$. If it is split on the column side, the rectangles $(\tau ,{\sigma }_j)$ are disjoint and cover $\tau \times \sigma$ because the ${\sigma }_j$ partition $\sigma$. Thus every split replaces a block by a disjoint cover of that block.

Each split increases the depth of one cluster in a finite tree, so along every branch the sum of the remaining row and column depths decreases and the recursion terminates. Induct on this finite sum: a terminal node is its own terminal cover; a nonterminal node has children whose terminal covers partition their respective rectangles by the induction hypothesis, and step 1.1 shows that their union is a disjoint cover of the parent rectangle. Applying this to the root $(I,J)$ proves that the terminal rectangles partition $I\times J$.

Two distinct terminal rectangles lie in different children of their first common split, whose rectangles are disjoint by step 1.1. Hence all terminal rectangles are pairwise disjoint, completing the partition claim.