Source: Set Theory by Kenneth Kunen.
Lemma III.6.2: Let $\gamma$ be any limit ordinal, and assume that $\kappa:=cf(\gamma)>\omega$. Then the intersection of any family of fewer than $\kappa$ club subsets of $\gamma$ is club.
Proof: Fix clubs $C_{\xi}\subseteq\gamma$ for $\xi<\lambda$, and let $D=\underset{\xi}{\bigcap}C_{\xi}$. Every intersection of closed set is closed, but we must show that $D$ is unbounded, assuming that $\lambda<\kappa$. So, fixed $\delta_0<\gamma$; we shall prove that $\sup(D)>\delta_0$.
First, for $\xi<\lambda$, define $f_{\xi}: \gamma\to\gamma$ so that $f_{\xi}(\alpha)$ is the least ordinal in $C_{\xi}$ that is greater than $\alpha$. Then, by recursion on $n$, define $\delta_n<\gamma$ so that $\delta_{n+1}=\sup\{f_{\xi}(\delta_n): \xi<\lambda\}$; note that $\mathbf{\delta_{n+1}<\gamma}$ because $\mathbf{cf(\gamma)>\lambda}$.
My Question: I am stuck on the boldface text. I've shown that $\delta_{n+1}\subseteq\gamma$, therefore $\delta_{n+1}\leq\gamma$. But how does $cf(\gamma)>\lambda$ imply $\delta_{n+1}\neq\gamma$? Any help would be greatly appreciated.
Suppose for contradiction that $\delta_{n+1} = \gamma$. Then $\sup\{f_\xi(\delta_n):\xi<\lambda\} = \gamma$, so $X_{n+1} = \{f_\xi(\delta_n):\xi<\lambda\}$ is cofinal in $\gamma$. Since $|X_{n+1}|\leq \lambda$ (since the map $\xi\mapsto f_\xi(\delta_n)$ is a surjective function $\lambda\to X_{n+1}$), this implies $\mathrm{cf}(\gamma)\leq \lambda$, contradicting the hypothesis that $\lambda<\kappa = \mathrm{cf}(\gamma)$.