Prove if $C_{1}$ and $C_{2}$ are closed unbounded, then $C_{1} \cap C_{2}$ is closed unbounded.

I have been reading the book "Introduction to Set Theory" by Jech and Hrbacek and have come to an exercise that I'm having trouble with. The exercise is in the chapter on filters and ultrafilters in the section on closed unbounded and stationary sets.

The exercise has a statement before it :

Let $\kappa$ be a regular uncountable cardinal. A set $C \subseteq \kappa$ is closed unbounded if $\sup(C) = \kappa$ and $\sup(C \cap \alpha) \in C$ for all $\alpha < \kappa$.

Here I believe that $\alpha$ is an ordinal.

The exercise statement is :

If $C_{1}$ and $C_{2}$ are closed unbounded, then $C_{1} \cap C_{2}$ is closed unbounded.

My solution so far is :

So : \begin{equation} \sup(C_{1}) = \sup(C_{2}) = \kappa \tag{1} \label{1} \end{equation} and : \begin{equation} \alpha < \kappa \Rightarrow \sup(C_{1} \cap \alpha) \in C_{1} \text{ and } \sup(C_{2} \cap \alpha) \in C_{2} \tag{2}\label{2} \end{equation} We see that \eqref{2} implies : \begin{align} \alpha < \kappa & \Rightarrow \sup\left( \{ \gamma \in C_{1} \; \mid \; \gamma < \alpha \} \right) \in C_{1} \\ \alpha < \kappa & \Rightarrow \sup\left( \{ \gamma \in C_{2} \; \mid \; \gamma < \alpha \} \right) \in C_{2} \end{align} So any increasing sequence of elements (could be finite) of $C_{1}$ of length less than $\kappa$ has its supremum in $C_{1}$. Same for $C_{2}$. Any increasing sequence in $C_{1} \cap C_{2}$ is a subsequence of some sequence in $C_{1}$ and some sequence in $C_{2}$. So the limit (supremum) exists in $C_{1}$ and $C_{2}$. So it must exist in $C_{1} \cap C_{2}$. So : \begin{equation} \alpha < \kappa \Rightarrow \sup\left( \{ \gamma \in C_{1} \cap C_{2} \; \mid \; \gamma < \alpha \} \right) \in C_{1} \cap C_{2} \end{equation} and : \begin{equation} \alpha < \kappa \Rightarrow \sup\left( (C_{1} \cap C_{2}) \cap \alpha \right) \in C_{1} \cap C_{2} \; \checkmark \end{equation} and $C_{1} \cap C_{2}$ is closed. $\checkmark$

Now need to show : \begin{equation} \sup(C_{1} \cap C_{2}) = \kappa \end{equation} Not sure how to do this. Can someone help ?

Let $\alpha < \kappa$. By unboundedness of $C_1$ we find $\alpha_1 \in C_1$ with $\alpha < \alpha_1$. Then by unboundedness of $C_2$ there is $\alpha_2 \in C_2$ with $\alpha_1 < \alpha_2$. Repeating this process we find increasing $(\alpha_n)_{n < \omega}$ with $\alpha_n \in C_1$ for odd $n$ and $\alpha_n \in C_2$ for even $n$. Now let $\beta = \sup_{n < \omega} \alpha_n$, so by uncountability and regularity of $\kappa$ we have $\beta < \kappa$. Then also $\beta = \sup_{n \text{ odd}} \alpha_n$ and hence $\beta \in C_1$ by closedness. Similarly we find $\beta \in C_2$, so $\beta \in C_1 \cap C_2$ and $\alpha < \beta$. As $\alpha < \kappa$ was arbitrary we conclude that $\sup(C_1 \cap C_2) = \kappa$.