If there is an ineffablea subtle cardinal then such $\phi$ doesn't exist.
An ineffablesubtle cardinal is a cardinal $\kappa$ such that for every sequence of sets $\langle S_\alpha \mid \alpha < \kappa\rangle$ such that $S_\alpha \subseteq \alpha$ there isand a stationary setclub $A\subseteq \kappa$ such that for every$C\subseteq \kappa$ there are $\alpha < \beta$ in $A$,$C$ such that $S_\beta\cap \alpha = S_\alpha$.
In our case, let us fix some bijection $f$ between $On \times \{0\} \cup \omega \times \{1\}$ and $On$ such that $f(n, 1) < \omega$ for all $n < \omega$ and $f(\beta, 0) = \beta$ for all $\beta \geq \omega$. Define: $$S_\alpha = \{f(\beta, 0) \mid \phi(\beta)\in\phi(\alpha)\}\cup \{f(n, 1) \mid n \in \phi(\alpha)\cap\omega\}$$
Note that $S_\alpha \subseteq \alpha$ for all $\alpha\geq \omega$. Clearly, $S_\alpha \subseteq S_\beta$ implies $\phi(\alpha) \subseteq \phi(\beta)$. Also, it is clear that for all $\alpha \geq \omega$, $S_\alpha \subseteq \alpha$. Assume that $\kappa$ is ineffable thensubtle. Then there are $\alpha < \beta$$\omega\leq \alpha < \beta$ such that $S_\alpha = S_\beta\cap \alpha$ and in particular, $S_\alpha \subseteq S_\beta$.
Remark: ineffablesubtle cardinals are consistent with $V=L$ (assuming that they are consistent with ZFC) and if $0^\#$ exists then every Silver indiscernible is ineffablesubtle.