Proving that compact Hausdorff groups are cofiltered limits of compact Lie groups

Question

What is the easiest way to show that a compact hausdorff topological group is a closed subgroup of a product of finite dimensional Lie groups?

Here are the relevant definitions:

Definition: (compact hausdorff group) A compact hausdorff group is an internal group in the category of topological spaces.

Definition: (Lie group) A Lie group is an internal group in the category of smooth manifolds. Any such manifold is homeomorphic to a smooth one.

Definition: (Pro category) The pro-category of a category C is the full subcategory of the completion [C,Set]${}^{op}$ consisting of those objects which are a cofiltered limit of objects in C.

A cofiltered limit of compact Lie groups is a closed subset of a product, and vice versa. Here are a few thoughts:

(1) If the only clopen subgroups of a compact hausdorff group $G$ are $\{ e \}$ and $G$, then must $G$ be a Lie group?

(2) The Peter Weyl theorem shows that a compact Lie group is a closed subgroup of U(n), so that from the above, any compact group is a closed subgroup of $\Pi_{X}$ U(∞) for some set $X$.

Lemma 2:

Answer 1

This is basically equivalent to the Peter–Weyl theorem. As abx says in the comments, the Peter–Weyl theorem implies that the map

$$G \to \prod_i U(V_i)$$

where $V_i$ runs over all the f.d. irreducible unitary representations of $G$ is injective, so it presents $G$ as a closed subgroup of a product of compact Lie groups as desired.

A priori it's not clear that there exists any nontrivial such representation, and for non-compact groups it can happen that there aren't any. So one has to somehow use compactness to show that there are enough such representations to separate points of $G$. The Peter–Weyl theorem does this by finding them in $L^2(G)$, which is defined using Haar measure on $G$, and showing that $L^2(G)$ decomposes into a Hilbert space direct sum of f.d. irreducibles. Since the action of $G$ on $L^2(G)$ separates points, this implies that the action of $G$ on its f.d. unitary irreducibles separates points. This is all to say that, as far as I know, it is really necessary to prove the existence of Haar measure and show that $L^2(G)$ decomposes into f.d. irreducibles to get this result; otherwise, again, it's not clear a priori that a completely arbitrary compact Hausdorff group $G$ has any nontrivial f.d. representations at all.

If the only clopen subgroups of a compact hausdorff group $G$ are $\{ e \}$ and $G$, then must $G$ be a Lie group?

No; for example $G$ could be a solenoid, which is an inverse limit of copies of $S^1$ (and hence connected). Solenoids are the groups Pontryagin dual to subgroups of $\mathbb{Q}$, for example the localizations $\mathbb{Z} \left[ \frac{1}{p} \right]$.