= Solution
{c}
Let $s_p$ be the finite stem of a condition $p=(n_p,s_p,A_p)$, and define
$$
f=\bigcup_{p\in G}s_p.
$$
The stems of two conditions in the <generic filter> agree on their common domain, because they have a common stronger extension. Hence this union is a function. For every $\ell\in\omega$, the set of conditions with $n_p\geq\ell$ is dense: append values from the nonempty infinite reservoir until the desired length is reached. Genericity therefore makes $f$ total on $\omega$. The generic stems, and thus their union, are available in $M[G]$.
Fix $g\in M\cap\omega^\omega$ and $K\in\omega$. In $M$ form the set
$$
D_{g,K}=\{(n,s,A):\exists k\ (K\leq k<n\ \land\ g(k)<s(k))\}.
$$
We show that it is a <dense subset of a forcing order>. Given $p=(n,s,A)$, put $k=\max(n,K)$. Fill the new positions $n,\ldots,k-1$ with any fixed element of $A$, and choose $a\in A$ with $a>g(k)$ for the new position $k$. This is possible because an infinite subset of $\omega$ is unbounded. Let $t$ be the resulting stem of length $k+1$, and keep the reservoir $A$ unchanged. Then
$$
q=(k+1,t,A)\leq p,\qquad q\in D_{g,K}.
$$
Every new stem value came from the old reservoir, exactly as required by the extension relation. The construction is performed in $M$, so $D_{g,K}\in M$ and is internally dense.
Genericity gives a condition in $G\cap D_{g,K}$. Its witnessing coordinate remains fixed in every later stem and hence in $f$, so some $k\geq K$ satisfies $g(k)<f(k)$. Since this holds for every $K$, there are infinitely many such $k$. As $g$ was arbitrary,
$$
\boxed{f\in\omega^\omega\cap M[G],\qquad\forall g\in M\cap\omega^\omega\;\exists^\infty k\;(g(k)<f(k)).}
$$
Thus this <infinite-reservoir stem forcing> produces an <unbounded real over a model>, which is precisely the stipulated meaning of bounding $M$.
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