= Solution
Conditions in $G$ are compatible finite functions, so their union $F=\bigcup G$ is a function $(\omega_1)^M\rightharpoonup(\omega_1)^M$. For each $\alpha<\omega_1$, the set
$$
D_\alpha=\{p:\alpha\in\operatorname{dom}p\}
$$
is dense: choose a normal function extending $p$ and add its value at $\alpha$. Genericity makes $F$ total.
If $\alpha<\beta$, choose a condition in the filter extending conditions that decide both values. It is contained in a <normal function on an ordinal>, so $F(\alpha)<F(\beta)$. Thus $F$ is strictly increasing.
It remains to prove continuity. For every limit $\delta<\omega_1$ and $\gamma<\omega_1$, let $D_{\delta,\gamma}$ contain the conditions $p$ such that $\delta\in\operatorname{dom}p$ and either
$$
p(\delta)\le\gamma
$$
or there is some $\alpha<\delta$ in $\operatorname{dom}p$ with $p(\alpha)>\gamma$. This set is dense. Given $p$, extend it to a normal function $f$ and add $(\delta,f(\delta))$; if $f(\delta)>\gamma$, continuity of $f$ supplies an $\alpha<\delta$ with $f(\alpha)>\gamma$, which may also be added.
Now fix $\gamma<F(\delta)$. Since $G$ meets $D_{\delta,\gamma}$, compatibility with the condition deciding $F(\delta)$ rules out the first alternative and gives $\alpha<\delta$ with $F(\alpha)>\gamma$. Therefore values below $\delta$ are cofinal in $F(\delta)$. Strict increase supplies the reverse bound, so
$$
\boxed{F(\delta)=\sup_{\alpha<\delta}F(\alpha).}
$$
Hence $F$ is normal on $(\omega_1)^M$ in $M[G]$.
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