= Solution
There is a definitional qualification. If a <compactly generated group> means a group algebraically generated by a compact subset, the assertion as printed is false. In the usual topology on the additive group $\mathbb Q$, the set
$$
K=\{0\}\cup\{1/n!:n\geq1\}
$$
is compact and generates $\mathbb Q$: every positive integer denominator divides some factorial. The group is metrizable, but it has no nonzero <Haar measure> finite on compact sets. Indeed, translation invariance gives a common singleton mass $a\geq0$. If $a>0$, the infinite compact set $K$ has infinite mass; if $a=0$, countable additivity gives zero mass to the whole countable group. This proves that <compact generation does not imply local compactness> or existence of <Haar measure>.
If the convention includes a compact generating neighbourhood, local compactness is already part of the hypothesis. More explicitly, the standard corrected result is \b[every locally compact, compactly generated metrizable group has <Haar measure>]. Here is a construction, rather than an invocation of that existence theorem.
Fix a nonzero $h\geq0$ in the <space of continuous compactly supported functions> $C_c(G)$. For nonzero $u\geq0$ in $C_c(G)$ and $f\geq0$ in $C_c(G)$, define the covering number and its normalized functional by
$$
(f:u)=\inf\left\{\sum_{j=1}^r a_j:f(t)\leq\sum_{j=1}^r a_j u(x_j^{-1}t)\ \text{for all }t,\ a_j>0\right\},\qquad I_u(f)=\frac{(f:u)}{(h:u)}.
$$
These numbers are finite: translates of the region where $u>0$ cover the <compact support> of $f$, and a finite subcover, multiplied by a sufficiently large coefficient, dominates $f$. They are positive when $f\neq0$, since every cover has total coefficient at least $\|f\|_\infty/\|u\|_\infty$. Composition of covers gives $(f:u)\leq(f:h)(h:u)$. Thus, with $M_f=(f:h)$,
$$
0\leq I_u(f)\leq M_f,\qquad I_u(h)=1.
$$
The functionals are homogeneous, monotone, subadditive and exactly invariant under left translation; the last property follows by relabelling the covering translates.
The key step in <Haar measure from normalized covering functionals> is asymptotic additivity as $\operatorname{supp}u$ shrinks to $e$. Given $f_1,f_2\geq0$, put $f=f_1+f_2$ and choose $p\in C_c(G)$ with $p\geq1$ on both supports. For $\delta>0$, set $\alpha_i=f_i/(f+\delta p)$, extended by zero where the denominator vanishes. These are continuous <compactly supported> functions and $\alpha_1+\alpha_2\leq1$. For any $\eta>0$, a sufficiently small identity neighbourhood $V$ ensures
$$
|\alpha_i(xv)-\alpha_i(x)|<\eta\quad\text{for every }x\in G,\ v\in V.
$$
Take $\operatorname{supp}u\subseteq V$ and any cover $f\leq\sum_j a_jT_{x_j}u$, where $T_xu(t)=u(x^{-1}t)$. Since $f_i\leq\alpha_i f+\delta p$, it follows that
$$
(f_i:u)\leq\sum_j a_j(\alpha_i(x_j)+\eta)+\delta(p:u).
$$
Adding and then taking the infimum over the cover of $f$ gives
$$
0\leq I_u(f_1)+I_u(f_2)-I_u(f)\leq2\eta M_f+2\delta M_p.
$$
First make $\delta$ small, then $\eta$ small, and finally the support of $u$ small enough for the oscillation bound. The additivity defect therefore tends to zero.
Compact generation gives a countable compact cover by finite products of a compact symmetric generating set. Local compactness upgrades this to a compact exhaustion with interiors covering $G$. Metrizability supplies a countable identity-neighbourhood base and a countable family of nonnegative $C_c$ functions dense in the uniform norm on each fixed <compact support>; this is the countable family whose existence may be assumed. Choose nonzero $u_n\geq0$ with supports shrinking to $e$. The uniform bounds $I_{u_n}(f)\leq M_f$ and a diagonal subsequence give convergence on that countable family. To extend convergence to every nonnegative $f$ supported in a fixed compact $L$, choose a cutoff $p_L\geq1$ on $L$. Monotonicity and subadditivity give
$$
|I_u(f)-I_u(g)|\leq I_u(|f-g|)\leq\|f-g\|_\infty M_{p_L}
$$
for $f,g$ supported in $L$. Uniform approximation therefore makes the chosen subsequence Cauchy on every such $f$.
Let $I(f)$ be its limit. Asymptotic additivity, homogeneity and invariance give an additive positive invariant functional on the nonnegative cone, with $I(h)=1$. Extend it by differences to real $C_c(G)$ and then complex linearly. It is bounded on each fixed <compact support> by the cutoff bound. The <Riesz representation on compactly supported continuous functions> consequently produces a positive <Radon measure> $m$ with $I(f)=\int f\,dm$. Invariance of $I$ and uniqueness of that representation imply $m(xE)=m(E)$. The measure is nonzero because $I(h)=1$, and finite on compact sets by the cutoff estimate. Moreover, for any nonzero $f\geq0$, composition of covers gives $I_u(f)\geq1/(h:f)>0$, so every nonempty open set has positive measure. Hence
$$
\boxed{m\text{ is a nonzero left Haar measure on the locally compact group}.}
$$
This completes the intended construction under the convention supplying local compactness, while the rational example settles the weaker literal interpretation.
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