Pairs and means
Prove that for any set \{\mathbf{x}_i\}_{i=1}^N with mean \bar{\mathbf{x}},
\sum_{i=1}^N\sum_{l=1}^N\|\mathbf{x}_i-\mathbf{x}_l\|^2=2N\sum_{i=1}^N\|\mathbf{x}_i-\bar{\mathbf{x}}\|^2.
Use it to show that the pairwise within-cluster scatter equals \sum_k N_k\sum_{C(i)=k}\|\mathbf{x}_i-\bar{\mathbf{x}}_k\|^2.
Subtract the common mean inside each pairwise difference. The cross term vanishes only after summing, which is the step that converts pairwise scatter into scatter around a centre.
Add and subtract \bar{\mathbf{x}}:
\|\mathbf{x}_i-\mathbf{x}_l\|^2 =\|\mathbf{x}_i-\bar{\mathbf{x}}\|^2+\|\mathbf{x}_l-\bar{\mathbf{x}}\|^2 -2(\mathbf{x}_i-\bar{\mathbf{x}})^\top(\mathbf{x}_l-\bar{\mathbf{x}}).
Sum over all N^2 pairs. The first term sums to N\sum_i\|\mathbf{x}_i-\bar{\mathbf{x}}\|^2, because each i appears with N values of l. The second term sums to the same quantity. The cross term sums to
-2\Bigl(\sum_i(\mathbf{x}_i-\bar{\mathbf{x}})\Bigr)^\top\Bigl(\sum_l(\mathbf{x}_l-\bar{\mathbf{x}})\Bigr)=0.
So the total is 2N\sum_i\|\mathbf{x}_i-\bar{\mathbf{x}}\|^2. Apply this inside each cluster k, with N=N_k, and multiply by \tfrac12. The pairwise definition of W(C) becomes
W(C)=\sum_k N_k\sum_{C(i)=k}\|\mathbf{x}_i-\bar{\mathbf{x}}_k\|^2.