I think you can get a much faster (and maybe easier...) proof using Riemannian geometry, as follows: 

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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 easy, see, e.g., Thm 2.28 in [these notes][1]). The side we will use ($G$ compact semi-simple $\Rightarrow$ $B$ neg.-def.) actually follows directly from $B(X,X)=tr(ad(X)\cdot ad(X))$ using an orthonormal basis with respect to an auxiliary bi-invariant metric to compute this trace.

Now, the Ricci curvature of any bi-invariant metric on $G$ (that exists because $G$ is compact) can be computed as:
 $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same [notes][1]. By the observation above, since $G$ is compact and semi-simple, its Killing form $B$ is negative-definite. Hence the above formula gives $Ric>0$. So, by the Bonnet-Myers Theorem, $G$ must have finite fundamental group. Q.E.D.

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