Every compact Lie group is a closed matrix group

Choose $U$ as in [L2]. For every $x\in G\setminus U$ there is, by [L1], a finite-dimensional unitary representation ${\pi }_x$ with ${\pi }_x(x)\mathrm{\ne }\mathrm{i}\mathrm{d}$; by continuity of ${\pi }_x$ there is an open neighbourhood $V_x$ of $x$ on which ${\pi }_x$ is nontrivial (does not contain the identity value). The sets $\{V_x\}$ cover the compact set $G\setminus U$, so finitely many of them, say $V_{x_1},\dots ,V_{x_m}$, already cover it.

Let $\pi :={\pi }_{x_1}\oplus \cdots \oplus {\pi }_{x_m}$ be the block-diagonal direct sum of [L3], a finite-dimensional unitary representation. If $g\in \ker \pi$ then ${\pi }_{x_j}(g)=\mathrm{i}\mathrm{d}$ for all $j$; by the choice of the $V_{x_j}$ this forces $g\mathrm{\notin }G\setminus U$, so $g\in U$, and $\ker \pi$ is a subgroup contained in $U$, hence $\ker \pi =\{e\}$ by [L2]. Thus $\pi$ is faithful.

A faithful representation is an injective continuous homomorphism $\pi :G\to U(N)$, with $N={\sum }_j\dim {\pi }_{x_j}$; its domain is compact and $U(N)$ is Hausdorff, so $\pi$ is a homeomorphism onto its compact image $H:=\pi (G)$, which is therefore closed in $U(N)$. Give $H$ its unique embedded Lie-group structure by [L4]. The corestriction $\pi :G\to H$ is a continuous homomorphism and hence smooth by [L4]. Since it is a homeomorphism, $\dim G=\dim H$ by [L4]. If $X\in \ker (d{\pi }_e)$, naturality of the exponential gives $\pi (\exp (tX))=\exp (t,d{\pi }_{e}X)=e$ for every $t$; injectivity of $\pi$ then gives $\exp (tX)=e$, and differentiation at $t=0$ yields $X=0$. Thus $d{\pi }_e$ is injective and, by equality of dimensions, is an isomorphism. Translation makes $d\pi$ invertible everywhere, so [L4] gives smooth local inverses. They agree with the global set-theoretic inverse, proving that $\pi :G\to H$ is a Lie-group isomorphism. Hence $G$ is isomorphic to the closed matrix Lie group $H\le U(N)$.