If $C$ is positive semidefinite, then so is $\begin{bmatrix} C & C\\ C & C\end{bmatrix}$ for the simple reason that it is nothing but the Kronecker product of $\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$ with $C$.
EDIT
For those who don't like Kronecker products, here is an alternative proof using block-matrices:
Since $C>0$, we can write $C^{1/2}$. Then, we can write \begin{equation*} \begin{bmatrix} C & C\\ C & C\end{bmatrix} = \begin{bmatrix} C^{1/2} & 0\\ C^{1/2} & 0\end{bmatrix} \begin{bmatrix} C^{1/2} & C^{1/2}\\ 0 & 0\end{bmatrix}, \end{equation*} which is a Gram matrix, hence semidefinite.