Topologically simple compact topological groups In a lecture note, there is the following sentence without any reference or proofs:

Is there any reference or proofs for it?
 A: First you should prove, using the Peter-Weyl theorem (a proof of which is in Chapter 4 of Lie Groups: Beyond an Introduction by Knapp), that 
Lemma. Every compact topological group which is topologically simple is isomorphic to a compact Lie group, and is algebraic.
Sketch of Proof.  I will just explain how to obtain an injective homomorphism into a matrix group. It is not too hard to see from the theorem that every compact topological group is isomorphic to a subgroup of a (possibly infinite) product of compact Lie groups. (This can be seen by first proving every element $x$ in the group admits a finite-dimensional complex unitary representation $\rho_x$ such that $x$ is not in its kernel; then $\bigoplus_x \rho_x$ is an injective map into a product of compact Lie groups of the form $U_n$ for some $n$). 
Now, let $G$ be a compact topological group which is topologically simple; embedding it in a product of compact Lie groups, we see that the projection onto each coordinate is either trivial or injective (since its kernel is a closed normal subgroup). At least one must be injective, and therefore $G$ must be isomorphic to a compact Lie group. $\square$
Thus, we just need to prove a compact Lie group $G$ which is topologically simple is either finite or connected.
For this you need to use the fact the connected component of the identity is an open-closed normal subgroup. Therefore, it is either trivial or everything. If it is trivial our group is discrete (since the trivial group is open), and compact, hence finite. If it is everything, our group is connected.
Observe this is not a necessary conditions: there are compact topologically simple groups which are neither finite nor connected algebraic.
A: First of all, by a compact topological group one usually understands a topological group which, is a compact Hausdorff topological space. I think, Hausdorffness assumption here is due to Bourbaki who always assume  compact topological spaces to be Hausdorff (without Hausdorffness assumption they use the terminology quasicompact). With this in mind, you can find a proof of the fact that every topologically simple compact group is abstractly simple, for instance, in Theorem 9.90 of 
Karl H. Hofmann, Sidney A. Morris, 
The Structure of Compact Groups. 
A Primer for the Student – A Handbook for the Expert, 
de Gruyter Studies in Mathematics, Volume 25, 1998. 
Without Hausdorffness, the claim is clearly false since every group can be equipped with the trivial topology which makes it compact. Such a group will be topologically simple (since it contains only two closed subsets) but not necessarily algebraically simple. Consider, for instance, the additive group of integers. 
Edit. One can replace Hausdorff by T1 since a topological group is Hausdorff if and only if it is T1.
