Base Case :- When we have just root then, there are no non leaf nodes. So No of leaves $= 1$, No of non leaf nodes is $= 0$. Base case holds.
Induction Hypothesis :- Assume that now for $k$ internal nodes we will have $k+1$ leaves.
Inducting on no of leaves, Now we add $2$ more leaves to this tree. One of $k+1$ leaf will become internal node. So now we will have $k+1$ internal node. No of leafs will be $K+ 1 - 1$ ($1$ leaf just became internal node) $+ 2$(New leafs) . So we proved that for any binary tree, in which every non-leaf node has $2$-descendants, the number of leaves in the tree is one more than the number of non-leaf nodes.