I think you can get a much faster proof using basic Riemannian geometry (maybe this was the sketch of proof originally suggested to the OP?) as follows:

---

First, recall that a semi-simple connected Lie group $G$ is compact if and only if its Killing form $B$ is negative-definite (the proof is quite simple, see, e.g., Thm 2.28 in [these notes][1]).

Since you start with a compact semi-simple Lie group $G$, by compactness, it has a bi-invariant metric. The Ricci curvature of this metric can be computed as: $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same [notes][1]. By the first observation, since $G$ is compact, its Killing form $B$ is negative-definite. Hence the above formula gives $Ric>0$. Finally, by the Bonnet-Myers Theorem, $G$ must have finite fundamental group.


  [1]: http://arxiv.org/pdf/0901.2374.pdf