 Solution

Suppose a tree parent at depth $n-1$, with $n<L$, has an unused outgoing digit. The missing tree child gives an empty code tree branch at depth $n$. Move any deepest tree leaf into that position, retaining all other codewords and deleting the abandoned empty branch. The new word has no old word as a prefix, because it lies in an unused branch, and it is not a prefix of any retained word. Its length decreases from $L$ to $n$. With a full-support finite distribution this reduces expected codeword length by $p_i(L-n)>0$, contradicting optimality. Hence all three tree children exist, and each has a descendant codeword. In fact every internal vertex above depth $L-1$ has all three tree children. Applying this to the tree parent of the specified prefix gives the claimed three extensions.

The positive-probability qualification attached only to the specified prefix does not by itself suffice. In the zero-weight example $(p,c)=((1,0,0),(0,10,110))$, choose the positive-probability word $0$, so $n=1<L=3$. There is no word starting with $2$. The deepest tree leaf available for relocation has probability zero, so the strict-improvement argument fails. This is why the full-support finite distribution assumption is used for the structural conclusions.