I think you can do a much faster proof using basic Riemannian geometry as follows. Let $G$ be a semi-simple connected Lie group. Then $G$ is compact if and only if its Killing form $B$ is negative-definite (see, e.g., Thm 2.28 in [these notes][1]). Since $G$ is compact, it has a bi-invariant metric, and the Ricci curvature of this metric is $Ric(X,Y)=-\frac14 B(X,Y)$ (see Remark 2.27 in the same notes). Since $B$ is negative-definite, it follows that $Ric>0$, so by the Bonnet-Myers Theorem, $G$ must have finite fundamental group.


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