= Solution
We construct a normalized positive translation-invariant functional on $C(G)$ and then represent it by a measure. The main step is a finite <matching in a graph> argument, for which we establish all the covering-net properties.
Let $V$ be an open neighbourhood of the identity in the compact <topological group> $G$. The sets $xV$, $x\in G$, form an open cover. Compactness supplies a finite set $F$ with $G=FV$. Among all such finite sets choose one of smallest cardinality, denoted $m(V)$. This is a <minimal left covering net of a compact group>. If $g\in G$, then $G=gFV$, and $gF$ has the same cardinality. Hence $gF$ is also minimal for the same $V$.
Let $F,F'$ be any two <minimal left covering nets of a compact group> for $V$. They both have cardinality $m(V)$. Form a <bipartite graph> by joining $x\in F$ to $y\in F'$ whenever $xV\cap yV\ne\varnothing$. For a subset $S\subseteq F$, let $N(S)$ be its set of neighbours. Every point of $SV$ lies in some $yV$ with $y\in F'$; that $y$ is a neighbour of a member of $S$. Therefore
$$
SV\subseteq N(S)V,
$$
and $(F\setminus S)\cup N(S)$ is still a left covering set. By minimality,
$$
m(V)\leq |(F\setminus S)\cup N(S)|\leq m(V)-|S|+|N(S)|.
$$
Thus $|N(S)|\geq|S|$. <Hall's marriage theorem> gives a <bijection> $\pi:F\to F'$ with $xV\cap\pi(x)V\ne\varnothing$ for every $x$. If $xv=\pi(x)v'$ at an intersection point, then
$$
\boxed{x^{-1}\pi(x)=vv'^{-1}\in VV^{-1}.}
$$
This proves the <matching minimal covering nets of a compact group> property, including the estimate needed below.
For a continuous complex-valued function $f$ on $G$, define
$$
I_F(f)=\frac1{|F|}\sum_{x\in F}f(x),\qquad
\omega_f(W)=\sup\{|f(xw)-f(x)|:x\in G,\ w\in W\}.
$$
Continuity and compactness imply $\omega_f(W)\to0$ as identity neighbourhoods $W$ shrink. Here is the uniformity argument: continuity of $(x,w)\mapsto f(xw)-f(x)$ at each $(x,e)$ gives a product neighbourhood where its absolute value is small; finitely many of the first-coordinate neighbourhoods cover $G$, and intersecting the corresponding second-coordinate neighbourhoods gives one $W$ that works for every $x$.
Matched pairs therefore give
$$
|I_F(f)-I_{F'}(f)|\leq\omega_f(VV^{-1}).
$$
In particular, with $F'=gF$,
$$
\boxed{|I_F(f\circ L_g)-I_F(f)|\leq\omega_f(VV^{-1}),\qquad L_g(x)=gx.}
$$
The estimate holds for every $g\in G$; no commutativity of $G$ is being assumed.
Direct the open identity neighbourhoods by reverse inclusion and choose a <minimal left covering net of a compact group> $F_V$ for each. The values $I_{F_V}(f)$ lie in the closed complex disk of radius $\|f\|_\infty$. The <Tychonoff theorem> makes the product of these disks over all $f\in C(G)$ compact. Consequently the <net> of vectors $(I_{F_V}(f))_{f\in C(G)}$ has a convergent <subnet>. Call its coordinatewise limit $I(f)$. Since each finite average is linear, positive on nonnegative real functions, and equal to one at the constant function one, the limit satisfies
$$
I(\alpha f+\beta h)=\alpha I(f)+\beta I(h),\qquad I(f)\geq0\ (f\geq0),\qquad I(1)=1,
$$
and $|I(f)|\leq\|f\|_\infty$. Thus $I$ is a bounded <positive linear functional>.
Given an identity neighbourhood $W$, continuity of multiplication and inversion supplies a sufficiently small $V$ with $VV^{-1}\subseteq W$. The translation estimate consequently tends to zero along the <subnet>. For every $f\in C(G)$ and every $g\in G$,
$$
\boxed{I(f\circ L_g)=I(f).}
$$
The <Riesz-Markov-Kakutani representation theorem> now represents $I$ by a regular positive <Borel measure> $\mu$ with $\mu(G)=1$. The preceding equality and uniqueness in that representation imply that the pushforward under every left translation equals $\mu$. Hence
$$
\boxed{\mu(gE)=\mu(E)\quad\hbox{for every Borel set }E\subseteq G.}
$$
This is the required normalized <Haar measure>. It is nonzero because its total mass is one. In fact every nonempty open set has positive measure: its left translates cover $G$, a finite subcover exists, and a zero measure for that open set would force $\mu(G)=0$.
For completeness, the constructed probability measure is also right invariant. Apply <Fubini's theorem> to $f(x^{-1}y)$ with both variables distributed according to $\mu$. Integrating first in $y$ gives $\int f\,d\mu$ by left invariance. Integrating first in $x$ and substituting $x=yz$ gives $\int f(z^{-1})\,d\mu(z)$. Thus $\mu$ is invariant under inversion. Inversion turns a right translation into a left translation, so right invariance follows. This last observation is not needed for the existence of a left <Haar measure>, but confirms the usual two-sided normalization for <compact groups>.
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