Principal Component Analysis, Eigenvectors lying in the span of the observed data points?
The statement says, that (2) and (4) are equal. That means (2)$\Rightarrow$(4) and (4)$\Rightarrow$(2). The first implication is trivial, as you correctly pointed out.
$$\lambda v=Cv$$
implies
$$\lambda x^Tv=x^TCv$$
for all $v$, not just those from $D$. But the second implication is a bit more tricky, but that is what the proof is about. It tells you, that if you want to check, whether a vector is an eigenvector of $C$, you don't have to check if (2) is satisfied. It tells you, that when (4) is satisfied, (2) is satisfies as well.

Imagine you have 2 points in 3D space. Those 2 points are
$$x_1=(-1,-1,0)$$
$$x_2=(1,1,0)$$
(please excuse me for not making this "dataset" centered) Note, that both points lie in the $xy$ plane. Now the correlation matrix is
$$C=\begin{bmatrix}
       1/2 &amp; -1/2 &amp; 0           \\[0.3em]
       -1/2 &amp; 1/2           &amp; 0 \\[0.3em]
       0           &amp; 0 &amp; 0
     \end{bmatrix}$$
Now you want to know, whether $v=[1\ -1\ 0]^T$ is an eigenvector with eigenvalue 1. The statement tells you, that you can either just check, if (2) is satisfied (3 equations) or if
$$\frac{1}{2}x_1^Tv=x_1^TCv=0$$
and
$$\frac{1}{2}x_2^Tv=x_2^TCv=0$$
which are only two equations.