Currently I'm reading (parts of) this survey. Though it is probably an easy idea that I'm missing, I'm stuck at the "$\left(2\right)\Rightarrow\left(3\right)$" direction of Proposition 5.11. It makes use of the following statement:
Let $\Gamma$ be an infinite discrete group. Then the intersection of the reduced group $C^\ast$-algebra $C_{\lambda}^\ast(\Gamma) \subseteq \mathcal{B}(\ell^2(\Gamma))$ with the compact operators $\mathcal{K}(\ell ^2(\Gamma))$ is empty.
The argument goes as follows: Assume that the intersection is not empty. Then we can find a non-zero finite rank projection in $C_\lambda^\ast(\Gamma)$ and also such a projection in the group $C^\ast$-algebra $C_\rho^\ast(\Gamma)$ of the right regular representation. Therefore $\lambda$ (the left regular representation) would have a finite-dimensional invariant subspace, in contradiction to $\Gamma$ being infinite.
The only step that is not clear to me is the last one. Why is having a finite-dimensional invariant subspace of the left regular representation only possible for finite groups?
The left-regular representation of a group is $C_0$, in the sense that $\langle \lambda(g)v,w\rangle$ tends to $0$ when $g$ tends to infinity, for every fixed $v,w$. This is essentially immediate.
Now every subrepresentation of a $C_0$-representation is $C_0$.
Finally, for an infinite group, no nonzero finite-dimensional unitary representation is $C_0$. Indeed, the above condition can be restated as: $\lambda(g)v$ tends weakly to $0$ when $g\to\infty$. In finite dimension, this means $\lambda(g)v\to 0$ for every $v$, i.e., $\|v\|=\|\lambda(g)v\|\to 0$ for every $v$. This can happen only if the Hilbert space is reduced to $\{0\}$.