 Solution

Let $T$ be ZFC together with the assertion of unboundedly many inaccessible cardinals. Given a model of $T$, if it has no inaccessible limit of inaccessibles, it already witnesses the required nonimplication. Otherwise let $\iota$ be its least inaccessible limit of inaccessibles and pass to its rank segment $V_{\iota }$.

Because $\iota$ is an inaccessible cardinal, $V_{\iota }$ satisfies ZFC. Its ordinals are precisely those below $\iota$, and below $\iota$ the inaccessible cardinals are unbounded by the choice of $\iota$. Inaccessibility of any smaller ordinal is absolute to this segment: it has the same subsets and functions on all smaller ordinals, as in strong-inaccessibility absoluteness from rank agreement. Thus $V_{\iota }$ satisfies the unbounded-inaccessibles assertion.

No smaller ordinal can be an inaccessible limit of inaccessibles, by minimality of $\iota$ and the same absoluteness. Also $\iota$ is not an ordinal of $V_{\iota }$ itself. ConsequentlyThis rank cutoff at the first inaccessible limit of inaccessibles proves $\mathrm{Con}(T)\Rightarrow \mathrm{Con}(T+\neg \mathrm{ILI})$. The soundness theorem for first-order logic then gives $T/⊢\mathrm{ILI}$ if $T$ is consistent. This is a model-theoretic relative-consistency argument, and does not infer external transitivity from mere consistency.