Proof with induction even number of letter
The proof is by induction on the length of $w$. If $|w| \leq 2$ then necessarily $w = a$ or $w = b$, and in both cases $w$ doesn't contain any $d$'s. If $|w| \geq 3$ then necessarily $w$ is of the form $ducvd$, where $u,v \in L$ are shorter words. By induction, each of $u,v$ contains an even number of $d$'s. It follows that $w$ also contains an even number of $d$'s: if $u,v$ contain $2s,2t$ many $d$'s (respectively), then $w$ contains $2s+2t+2=2(s+t+1)$, which is also even.