Suppose that $\gamma$ is a limit ordinal. We say that a subset $X \subseteq \gamma$ is closed if whenever $\delta < \gamma$ is a limit ordinal and $\sup(X \cap \delta) = \delta$, then $\delta \in X$. We say that $X \subseteq \gamma$ is unbounded if $\sup(X) = \gamma$. A set $X$ is club in $\gamma$ if $X$ is closed and unbounded in $\gamma$.
Now I am trying to show that if $\text{cf}(\gamma) = \omega$ (where $\text{cf}(\gamma)$ denotes the cofinality of $\gamma$), then any cofinal $\omega$-sequence is club in $\gamma$.
The text I am using has not skipped over cofinal sequences so am I correct to begin by letting $\langle \alpha_{\xi} \mid \xi < \omega \rangle$ be such that $\lim_{\xi \rightarrow \omega} \alpha_{\xi} = \gamma$ with the hopes of showing that this sequence is club in $\gamma$?
If so, I am unsure how to proceed from there. It seems pretty clear that the supremum over this sequence is equal to $\gamma$ and is therefore unbounded. The part I am having trouble with is how to show that it is closed. If I have some limit ordinal $\delta < \gamma$, I am unsure how to interpret the intersection of $\delta$ and this sequence. Do I need to show that for all $\xi < \omega$, if $\delta < \gamma$ and $\sup(\alpha_{\xi} \cap \delta) = \delta$ then $\delta$ belongs to this sequence?