= Solution
It suffices to obtain a <model of ZFC without weakly inaccessible cardinals>. Starting with any model of <ZFC>, pass to its <constructible universe>, which satisfies <ZFC> and the <Generalized continuum hypothesis>. If it has no <inaccessible cardinal>, use that model. Otherwise pass to its rank segment at its least <inaccessible cardinal> $\delta$. This segment satisfies <ZFC>, retains the <Generalized continuum hypothesis>, and has no <inaccessible cardinals>. Under the <Generalized continuum hypothesis>, every <weakly inaccessible cardinal> is strongly inaccessible: if $\lambda<\kappa$ and $\kappa$ is a <limit cardinal>, then $2^\lambda=\lambda^+<\kappa$. Thus in either case the resulting model $N$ has no <weakly inaccessible cardinals>.
Work inside $N$. If $\alpha$ is a nonzero <limit ordinal>, let $\lambda=\sup_{\beta<\alpha}\sigma_\beta$. It is an uncountable <limit cardinal>. If it were regular, it would be a <weakly inaccessible cardinal>, which is impossible in $N$. It is therefore singular, and the <singular cardinal enumeration> is continuous at this index:
$$
\sigma_\alpha=\sup_{\beta<\alpha}\sigma_\beta.
$$
The <cofinality of an increasing ordinal supremum> now gives
$$
\boxed{\operatorname{cf}(\sigma_\alpha)=\operatorname{cf}(\alpha)}
$$
for every nonzero <limit ordinal> $\alpha$ in $N$. Hence $N$ satisfies the negation of the proposed existential assertion. By the <soundness theorem for first-order logic>, consistency of <ZFC> prevents <ZFC> from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a <countable transitive model>. Here, as usual, a <limit ordinal> excludes zero.
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