= Solution
There is an essential qualification to the printed hypothesis. If <compactly generated group> means merely generated algebraically by a <compact subset>, the claim is false. In the additive group $\mathbb Q$ with its usual <topology>,
$$
K=\{0\}\cup\{1/n!:n\ge1\}
$$
is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant <Radon measure> has a common mass $c$ on each singleton. If $c>0$, the <compact set> $K$ has infinite measure, contradicting finiteness on <compact sets>. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero <Haar measure>. This proves that <compact generation does not imply local compactness> and disproves the literal unrestricted statement.
The intended existence statement is valid when local <compactness> is included, either explicitly or in the convention that the group has a compact generating <neighbourhood>. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.
Enlarge a compact generating set to a symmetric compact <neighbourhood> of the <identity element> $S$. Then $G=\bigcup_nS^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open <neighbourhood> of the <identity element> contained in $S$ supplies an open <neighbourhood> of every point of $S^n$. This gives a compact exhaustion. Every compact <metric> space has a countable <dense subset>: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the <metric> space is separable and has a countable open basis.
We use nonnegative <compactly supported> <continuous functions> and cutoffs equal to one on specified <compact sets>. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge0$ define the weighted covering quantity
$$
(f:u)=\inf\left\{\sum_{j=1}^m c_j:f\le\sum_{j=1}^m c_jT_{x_j}u,\ c_j\ge0\right\},\qquad
T_xu(y)=u(x^{-1}y),\qquad I_u(f)=\frac{(f:u)}{(h:u)}.
$$
The quantities are finite: an <open set> where $u$ is bounded below has translates covering the <compact support> of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\|u\|_\infty>0$.
Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ gives
$$
(f:u)\le(f:h)(h:u),\qquad 0\le I_u(f)\le(f:h).
$$
This is a bound independent of $u$. On functions supported in a fixed <compact set> $C$, choose a cutoff $k\ge1$ on $C$. Monotonicity and subadditivity imply
$$
|I_u(f)-I_u(g)|\le\|f-g\|_\infty I_u(k)\le\|f-g\|_\infty(k:h).
$$
Thus the covering functionals are uniformly Lipschitz on each fixed support.
Choose a nonzero $u_n\ge0$ with support in an <neighbourhood> of the <identity element> shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable <metric> cone in the <supremum norm>, as a subspace of <continuous functions> on that compact <metric> space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.
The crucial remaining step is additivity. Fix $f_1,f_2\ge0$ with combined support $C$, and fix a compact <neighbourhood> of the <identity element> $V_0$. Choose a cutoff $k$ equal to one on $CV_0^{-1}V_0$. For $\delta>0$, put
$$
F=f_1+f_2+\delta k,\qquad r_i=f_i/F,
$$
setting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a <neighbourhood> of the supports, and off those supports the numerator vanishes. Their sum is at most one. <Compactness> and continuity give, for each $\eta>0$, an <neighbourhood> of the <identity element> $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed <compact set> $CV_0^{-1}V_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that <compact set> give a single small right-increment <neighbourhood>, then choose $V^{-1}V$ inside it.
If $u$ has support in $V$ and $F\le\sum_j c_jT_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le\sum_j a_{ij}c_jT_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le1+2\eta$. Taking covering infima yields
$$
I_u(f_1)+I_u(f_2)\le(1+2\eta)I_u(F)
\le(1+2\eta)[I_u(f_1+f_2)+\delta I_u(k)].
$$
Along the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta\downarrow0$ and $\delta\downarrow0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.
Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The <Riesz representation on compactly supported continuous functions> gives a <Radon measure> $\mu$ with $I(f)=\int f\,d\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu(xE)=\mu(E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on <compact sets> because a cutoff dominating their indicators has finite integral. Finally every nonempty <open set> has positive measure: otherwise its translates cover the <compact support> of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence \b[$\mu$ is a <Haar measure> under the stated local-compactness qualification]. This is <Haar measure from normalized covering functionals>, with the limiting additivity proved explicitly.
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