 Solution

Closure preserves the ground-model cardinal $\lambda$. We use the stronger fact that closed forcing adds no short ordinal sequences. Fix an ordinal $\mu <\lambda$ in $M$ and a forcing name $\dot{h}$ that a condition $p$ forces to be an ordinal-valued function with domain $\mu$.

Inside $M$, recursively strengthen $p$ to decide each value $\dot{h}(\xi )$, for $\xi <\mu$. At a limit stage take a lower bound of the previously chosen descending sequence; its length is below $\lambda$, so $\lambda$-closure supplies one. After all $\mu$ values have been decided, closure supplies a common lower bound $q$. The decided values form a function $h\in M$ by the Axiom schema of replacement, andThe recursion is carried out within $M$ using the definable forcing theorem and a ground-model choice of deciding extensions, so every sequence to which closure is applied belongs to $M$. Repeating the construction below any stronger condition shows that such full-decision conditions are dense below a forcing condition $p$. The dense-below generic meeting lemma ensures that a generic filter containing $p$ meets them. Thus the interpreted function is in $M$.

If $\lambda$ ceased to be a cardinal number in $M[G]$, there would be a surjection $\mu \to \lambda$ for some ordinal $\mu <\lambda$. The forcing theorem gives a name and a condition in $G$ forcing this. The argument above makes its interpretation a ground-model function, but $M$ regards $\lambda$ as a cardinal number and admits no such surjection. ThereforeNo regularity of $\lambda$ is required; every recursion used has length strictly below $\lambda$. Finite cardinals are preserved automatically. This is cardinal preservation by closed forcing.