Assuming the dilation is non-identity (r$\neq $ 1).
It's a straight forward proof to show that $C$ is the only invariant point under the dilation, using the equation $C'=$ $D_C,r$ $(C)$.
$C=$($c_1$, $c_2$, $1$) and $D_C,r$ = $\begin{bmatrix}r&0&c_1\left(1-r\right)\\ \\0&r&c_2\left(1-r\right)\\ \\0&0&1\end{bmatrix}$ .
But I can't come up with a method to show that all lines incident with center $C$ are invariant.
It depends a lot on how you define a line.
You can define a line as the set of points $ax+by+d=0$. Then the homogeneous coordinates $[a:b:d]$ describe the line. Homogeneous here means that multiples of these numbers represent the same set of points. $3ax+3by+3d=0$ is the same line.
How does your matrix act on a line? A point lies on the transformed line if it is the image of a point on the original line. As a formula, you have the original point lying on the original line as
$$[a,b,d]\cdot\begin{bmatrix}x\\y\\1\end{bmatrix}=0$$
and then the transformed point lies on the transformed line
$$\left([a,b,d]\cdot D_{C,r}^{-1}\right)\cdot\left(D_{C,r}\cdot\begin{bmatrix}x\\y\\1\end{bmatrix}\right)=0$$
So if you represent the line as a row vector, you multiply that with the inverse of your transformation matrix from the right. If you treat your line as a column vector and use dot product between that and the point, then you multiply the inverse transpose transformation matrix from the left.
So where does all this lead? You are looking for a vector which, when multiplied with a matrix, changes perhaps its magnitude but not its direction. That's an eigenvector problem. You want to show that all lines through $C$ are eigenvectors of the inverse (and possibly transpose) matrix, belonging to an eigenspace of dimension two. (You would also find the line at infinity as another eigenvector of any affine transformation.)
If you represent your lines in a different way, say using a pair of distinct points, or point and direction, then your approach would likely be different. I'll leave this for other answers to cover.