= Solution
Give $C(G)$, the <space of continuous functions on a compact space>, its <uniform norm>. The <left regular action> is $l_gf(x)=f(g^{-1}x)$. The group law gives $l_gl_h=l_{gh}$ and $l_e=I$, and each translation is a linear <isometry> because $x\mapsto g^{-1}x$ is a bijection of $G$.
For fixed $f\in C(G)$, the map $(g,x)\mapsto f(g^{-1}x)$ is continuous. Compactness of the $x$-space makes this dependence uniform: for a given $g_0$ and $\varepsilon>0$, continuity gives a <neighbourhood> of $g_0$ and a <neighbourhood> of each $x$ on which the difference from $f(g_0^{-1}x)$ is less than $\varepsilon$; a finite cover of $G$ permits one common <neighbourhood> of $g_0$. Hence $\|l_gf-l_{g_0}f\|_\infty\to0$. The estimate
$$
\|l_gf-l_{g_0}f_0\|_\infty\leq\|f-f_0\|_\infty+\|l_gf_0-l_{g_0}f_0\|_\infty
$$
then proves the joint continuity of the action on $G\times C(G)$.
For a left action on the <continuous dual space>, use the contragredient convention
$$
T_g\phi=\phi\circ l_{g^{-1}}.
$$
It satisfies $T_gT_h=T_{gh}$ and preserves the dual <unit ball> $B'$. If the printed prime denotes the literal transpose $l_g^*\phi=\phi\circ l_g$, these maps instead satisfy $l_g^*l_h^*=l_{hg}^*$ and form a right action; replacing $g$ by $g^{-1}$ gives the left action above. Their invariant functionals are the same.
For $g_i\to g$ and $\phi_i\to\phi$ in the <weak-star topology> on $B'$, every $f\in C(G)$ satisfies
$$
|(T_{g_i}\phi_i)(f)-(T_g\phi)(f)|\leq\|l_{g_i^{-1}}f-l_{g^{-1}}f\|_\infty+|(\phi_i-\phi)(l_{g^{-1}}f)|\longrightarrow0.
$$
The first term uses strong continuity of translation and $\|\phi_i\|\leq1$; the second uses weak-star convergence at one fixed function. This proves the requested joint <weak-star topology> continuity.
To obtain <Haar measure>, consider the set $K$ of <positive linear functionals> $\phi$ on $C(G)$ with $\phi(1)=1$. Such functionals have <norm> one: for complex $f$, choose a phase making $\phi(f)$ real and nonnegative and use $\operatorname{Re}(e^{it}f)\leq\|f\|_\infty$; this gives $|\phi(f)|\leq\|f\|_\infty$, with equality on the constant-one function. Thus $K\subseteq B'$. Evaluation at the identity belongs to $K$, and positivity and normalization are weak-star closed conditions. The <Banach-Alaoglu theorem> makes $K$ a nonempty compact convex set. Each $T_g$ is a continuous affine self-map of it.
Here is the fixed-point step, including why no <commutativity> assumption on $G$ is needed. For any finite list $g_1,\ldots,g_n$, the <Schauder-Tychonoff fixed-point theorem> gives a fixed point $\phi\in K$ of $n^{-1}\sum_jT_{g_j}$. Let $H$ be the closed subgroup they generate and, for real $f\in C(G)$, set $h(x)=\phi(l_xf)$ on $H$. It is continuous and satisfies
$$
h(x)=\frac1n\sum_{j=1}^nh(g_j^{-1}x).
$$
At a maximum point $x_0$, every summand must have the same maximum value. Iteration gives that value at all inverse-generator words times $x_0$. The closure of their positive-word monoid is a <closed subsemigroup of a compact group>, hence is the entire generated subgroup $H$. Density and continuity make $h$ constant on $H$. Therefore $\phi$ is fixed by each of $T_{g_1},\ldots,T_{g_n}$, first on real functions and then on complex functions by linearity. The closed fixed-point sets consequently have the <finite intersection property> in $K$, so compactness gives a common fixed functional for every $g\in G$.
The <Riesz-Markov-Kakutani representation theorem> converts this positive functional into a regular Borel <probability measure> $\mu_l$ with $\phi(f)=\int f\,d\mu_l$. Its invariance gives left translation invariance of the <measure>. It has full support: if a nonempty open set had <measure> zero, a finite collection of its left translates would cover the compact group and force $\mu_l(G)=0$, contradicting normalization. Thus it is a normalized left <Haar measure>. The right translation action gives a normalized right <Haar measure> $\mu_r$ by the same argument.
For uniqueness and equality, take any $f\in C(G)$ and apply the <Fubini theorem> to
$$
I=\int_G\int_G f(yx)\,d\mu_l(x)\,d\mu_r(y).
$$
For fixed $y$, left invariance of $\mu_l$ makes the inner integral $\int f\,d\mu_l$. Reversing the order, for fixed $x$, right invariance of $\mu_r$ makes the inner integral $\int f\,d\mu_r$. Since both total masses are one,
$$
\int f\,d\mu_l=I=\int f\,d\mu_r.
$$
Uniqueness in the <Riesz-Markov-Kakutani representation theorem> gives $\boxed{\mu_l=\mu_r}$. Comparing any normalized left <Haar measure> to this fixed right <Haar measure> proves uniqueness as well. \b[The conclusion is uniqueness of normalized <Haar measure>.] Without normalization, the printed uniqueness claim is false: even on the one-point group, $\delta_e$ and $2\delta_e$ are distinct Haar <measures>. In general the result is uniqueness up to a positive scalar, and left and right probability normalizations coincide.
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