= Solution
To prove <replacement failure at V omega plus omega>, put $M=V_{\omega+\omega}$. The <set> $\omega$ belongs to $M$, and for every $n\in\omega$ the <ordinal> $\omega+n$ also belongs to $M$. The relation saying that $y$ is obtained by taking $n$ successive <ordinal> successors of $\omega$ is definable, with parameter $\omega$, and is single valued in $M$.
For clarity this does not assume <Replacement> inside $M$. For each finite $n$ its finite computation <sequence> exists in $M$: its entries have rank below $\omega+n+1$, and the finite coding overhead still has rank below $\omega+\omega$. The <bounded formula in set theory>[bounded formula] verifying this finite successor chain is absolute between $M$ and the universe. The ordinary finite induction identifying its unique last entry uses the actual finite <ordinals>, which are the same in $M$. This justifies the internal definable <function> $n\mapsto\omega+n$ without invoking the axiom being tested.
If <Replacement> held there, its range $B=\{\omega+n:n\in\omega\}$ would be an element of $M$. But <ordinal> rank equals the <ordinal>, so
$$
\operatorname{rank}(B)=\sup_{n<\omega}(\omega+n+1)=\omega+\omega.
$$
Every element of $V_{\omega+\omega}$ has strictly smaller rank. This contradiction proves \b[<Replacement> fails in $V_{\omega+\omega}$]. Each individual output is present; the missing object is the <set> collecting all outputs.
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