This is again a question about forcing. Start in $L$, the constructible universe. CH holds. Let $\lambda$ be an inaccessible cardinal, also let $\lambda$ > $\aleph_0$. For each $\alpha < \lambda$, let $P_\alpha$ be the set of all functions such that $dom(p_\alpha) \subset \aleph_0$, $|dom(p_\alpha)|<\aleph_0$ and $ran(p_\alpha) \subset \alpha$. $p_\alpha$ is stronger than $q_\alpha$ iff $p_\alpha$ extends $q_\alpha$.

Now consider $(P,<)$ be the $\kappa$-product of the $P_\alpha$, $\alpha<\lambda$. Now the conditions of $P$ are functions taking their argument on the set of all subsets of $\lambda \cdot \aleph_0$ such that the cardinality of the domain of $p$ is strictly smaller than $\aleph_0$ and such that $p(\alpha,\xi)<\alpha$ for each pair $(\alpha,\xi) \in dom(p)$.

If $G$ is a generic set of conditions then let for each $\alpha$, $G_\alpha$ be the projection of $G$ on each $P_\alpha$, each $G_\alpha$ is a generic filter, let $f_\alpha= \bigcup G_\alpha$ is a function fro, $\aleph_0$ onto $\alpha$ for every $\alpha < \lambda$ we have $|\alpha| \leq \aleph_0$.

Since the forcing is $<\aleph_0$-closed so cardinals and cofinalities are preserved and it satisfies the $\lambda$-chain condition so $\lambda$ is a cardinal in the generic extension $L[G]$.

Now we have $\lambda=\aleph_0^+$. But in light of the basic fact (that I had overlooked in my previous post), the continuum can't be strong limit. So the above is clearly false. Can you help me point my mistakes?