I know that the matrix $D$ is a real orthogonal matrix.
$D = \begin{bmatrix} B & -C\\ C & B \end{bmatrix}$
That would mean that $DD^T = I => \begin{bmatrix} B & -C\\ C & B \end{bmatrix} \begin{bmatrix} B & C\\ -C & B \end{bmatrix} = \begin{bmatrix} I & 0\\ 0 & I \end{bmatrix} => B^2 + C^2 = I$ and $BC - CB = 0 => BC = CB$
How would I use this(if I even need it) to prove that $B + iC$ is unitary?
Your transpose is not right. From $DD^T=I$ you get $$ \begin{bmatrix} B & -C\\ C & B \end{bmatrix} \begin{bmatrix} B^T & C^T\\ -C^T & B^T \end{bmatrix} = \begin{bmatrix} I & 0\\ 0 & I \end{bmatrix} ;$$ combined with $D^TD=I$, you get $$ BB^T+CC^T=B^TB+C^TC=I, \qquad BC^T-CB^T=C^TB-CB^T=0.$$ Now $$(B+iC)^*(B+iC)=(B^T-iC^T)(B+iC)=B^TB+C^TC+i(B^TC-C^TB)=I.$$ The computation for $(B+iC)(B+iC)^*$ is similar (and not needed since we are dealing with matrices).