Let $u$ and $v$ be unit vectors over some finite dimensional complex Hilbert space. We are given that
$$\|u u^{*} - v v^{*} \|_1 \leq \epsilon$$
We want to upper bound
$$\|u - v \|_2$$
I have tried using Fuchs van de Graaf and other inequalities but haven't succeeded much.
This is not possible. If $u = -v$ or $u=sv$ with $|s|=1$ then $uu^*-vv^*=0$ while $\|u-v\|_2 \ne0$ for $s\ne1$.
Now assume that $u,v$ are linearly independent. Since you are talking about Fuchs-van de Graaf, I suppose $\|T\|_1$ refers to the Schatten-1-norm of $T$, or the trace of the operator $(T^*T)^{1/2}$.
In our case $T = uu^*-vv^*$ is self-adjoint. So $\|T\|_1$ is the sum of the absolute values of the eigenvalues of $T$.
Now, the nontrivial eigenspaces of $T$ are spanned by $u$ and $v$, and the operator $T = uu^*-vv^*$ has the same non-zero eigenvalues as the matrix $\pmatrix{ 1 & -v^*u \\ u^*v & -1}$, which is the matrix of $T$ with respect to the basis $(u,v)$ of $span(u,v)$. These eigenvalues are $\pm \sqrt{1- |u^*v|^2}$. So $\|T\|_1 = 2\sqrt{1- |u^*v|^2}$. The vector difference can be written as $$\|u-v\|^2 = 2(1- \Re(u^*v)),$$ which seems structurally similar. But I cannot see how to bound it in terms of $\|T\|_1$ even for $T\ne0$.