The excercise is as follows
Let us say that $D\subset\kappa^{<\kappa}$ is dense if for any $s\in\kappa^{<\kappa}\exists t\in D$ such that $s\subset t$.
Prove that if $\kappa$ is regular, for every family of dense subsets $\{D_\alpha|\alpha<\kappa\}$ there is $x\in\kappa^\kappa$ so that $\forall\alpha<\kappa\exists s\in D_\alpha$ such that $s\subset x$.
I am having trouble visualizing the situation: intuitively I would build $x$ inductively, exploiting the fact that each $D_\alpha$ is dense. Anyhow I can't write a proper definition nor I see where $\kappa$ regular comes into play.
We can construct an increasing sequence $\langle s_\alpha\mid\alpha<\kappa\rangle$ over $\kappa^{<\kappa}$ such that $s_\alpha\in D_\alpha$ as follows: for $\alpha=0$, take any $s_0\in D_0$. If $\alpha=\beta+1$, take any $s_{\beta+1}\supseteq s_\beta$ such that $s_{\beta+1}\in D_{\beta+1}$. If $\alpha$ is a limit ordinal, choose $s_\alpha\supseteq \bigcup_{\xi<\alpha} s_\xi$ satisfying $s_\alpha\in D_\alpha$. (The assumption $\kappa$ is regular is necessary to ensure $\bigcup_{\xi<\alpha}s_\xi$ has length $<\kappa$.)
Then take $x=\bigcup_\alpha s_\alpha$: if $x$ is not a $\kappa$-sequence, make $x$ longer by putting any elements after $x$ so that $x\in \kappa^\kappa$. We can see that some initial segment of $x$ is a member of $D_\alpha$ by the construction.