Here is a direct construction of semigeneric conditions for Namba forcing (actually for Namba$^{\prime}$ for convenience) using $SCC$, without using games.

Let $\theta$ be large enough and regular, $X\prec H_\theta$ countable and $p\in X\cap Nm^\prime$. We construct a descending sequence $\langle p_n\mid n<\omega\rangle$ of condtions and make sure that the $n$th level remains constant after $n$ steps, so that we get a fusion $p_\omega$ in the end.
Let $\langle \dot \xi_n\mid n<\omega\rangle $ be an enumeration of all names for countable ordinals in $X$. Put $p_0=p$ and assume $p_n$ is constructed. Let $\eta$ be on the $n$th level of $p_n$. Find $r_\eta\leq p_n\upharpoonright \eta$ that has stem $\eta$ and decides as many of the $\dot\xi_0,\dots, \dot\xi_n$ as possible and let $\zeta_\eta$ be the supremum of the relevant decisions. We get $p_{n+1}$ by replacing all such subtrees $p_n\upharpoonright \eta$ by the corresponding $r_\eta$.

By making sure that all our choices are minimal with respect to some wellorder in $X$, we can make sure that all $p_n$ are in $X$ (even though $p_\omega$ is not). Even more, whenever $X\subseteq Y\prec H_\theta$ and $\eta\in p_\omega\cap Y$ then $\zeta_\eta\in Y$.

Now we will thin out $p_\omega$ (we need $SCC$ to make sure that we do not cut off too many branches). Let
$$q=\{\eta\in p_\omega\mid \exists Y\text{ countable with } X\sqsubseteq Y\prec H_\theta\wedge \eta\in Y\}$$
I claim that $q$ is semigeneric for $X$. First we have to see that $q$ is a condition. Suppose that $\eta\in q$ and that $Y$ witnesses this. Let $n=lh(\eta)$. Note that the direct successors of $\eta$ in $p_\omega$ are the same as the ones in $p_{n+1}\in Y$, so that we can find an enumeration $(\mu_\alpha)_{\alpha<\omega_2}\in Y$ of them. By $SSC$, there are unboundedly many $\beta<\omega_2$ for which there are countable $Y\sqsubseteq Z\prec H_\theta$ with $\beta\in Z$ an thus $\mu_\beta\in Z$. Hence every point in $q$ has $\omega_2$-many successors.

Finally suppose that $r\leq q$ decides $\dot \xi_n$ as $\xi$. Let $\eta$ be the stem of $r$, we may assume that $lh(\eta)\geq n$. By the construction of $p_{lh(\eta)+1}$, $\xi\leq\zeta_\eta$. Now let $Y$ witness that $\eta\in q$. We thus have $\xi\leq\zeta_\eta\in Y\cap\omega_1=X\cap\omega_1$ so that $\xi\in X$, as desired.