The answer is yes, and this question is strongly related to the topic of Hausdorff gaps.

You have specified a sequence of functions $x^{(k)}$ that is decreasing modulo finite the usual order on $\mathbb{R}^\omega/\text{Fin}$. That is, for each $k$, you have $x^{(k+1)}_n< x^{(k)}_n$ for all but finitely many $n$. Furthermore, if we let $z^{(k)}_n=k$ for all $n$ and $k$, we have what is called an $(\omega,\omega)$-gap, because for any $k$ we have $$z^{(k)}_n<z^{(k+1)}_n<\cdots<x^{(k+1)}_n<x^{(k)}_n$$ for all sufficiently large $n$. Hausdorff proved that in this situation, there is a function $n\mapsto y_n$ that fills the gap (and you can find explicit constructions of the gap-filling functions). So we have for each $k$ that $k<y_n<x^{(k+1)}_n$ for all but finitely many $n$. And this ensures your desired hypotheses.

The answer is yes, and this question is strongly related to the topic of Hausdorff gaps.

You have specified a sequence of functions $x^{(k)}$ that is decreasing modulo finite with respect to the coordinate-wise order on $\mathbb{R}^\omega/\text{Fin}$. We write $f<^*g$ to mean that $f(n)<g(n)$ for all but finitely many $n$. For each $k$, you have $x^{(k+1)}_n< x^{(k)}_n$ for all but finitely many $n$, and so $x^{(k+1)}<^*x^{(k)}$. Furthermore, if we let $z^{(k)}_n=k$ for all $n$ and $k$, we have what is called an $(\omega,\omega)$-(pre)gap, because for any $k$ we have $$z^{(k)}_n<z^{(k+1)}_n<\cdots<x^{(k+1)}_n<x^{(k)}_n$$ for all sufficiently large $n$. So we have an increasing sequence below converging upward to the gap and decreasing sequence above converging downward to the gap (using the order on functions modulo finite).

Hausdorff proved that in this situation, there is a function $n\mapsto y_n$ that fills the gap (and you can find explicit constructions of the gap-filling functions; Andreas provides one). (There are continuum many such functions, and no single optimal one.) So we have for each $k$ that $k<y_n<x^{(k+1)}_n$ for all but finitely many $n$. And this ensures your desired hypotheses.