I am trying to prove that $C^{T}DC$ is symmetric. I start by assuming that $B=C^{T}DC$. Using the rule of multiplication of matrices,
$$ b_{ij} = \sum \limits_{l} \sum \limits_{k} c_{il}^{T} d_{lk} c_{kj} = \sum \limits_{l} \sum\limits_{k} c_{li} d_{lk} c_{kj}$$
Similarly,
$$b_{ji} = \sum \limits_{l} \sum \limits_{k} c_{jl}^{T} d_{lk} c_{ki} = \sum \limits_{l} \sum \limits_{k} c_{lj} d_{lk} c_{ki} = \sum \limits_{l} \sum \limits_{k} c_{ki} d_{lk} c_{lj}$$
Since $k$ and $l$ are dummy indices, we can interchange them. (Right?)
This gives us
$$b_{ji} = \sum \limits_{l} \sum \limits_{k} c_{li} d_{kl} c_{kj} = b_{ij}$$
This proves that $C^{T}DC$ is always symmetric provided that $D$ is symmetric. Is this proof correct?
Yes, your proof is correct.
However, here is a much simpler proof: $$ B^T = (C^TDC)^T = C^TD^T(C^T)^T = C^TDC = B $$