The problem says:
For any $U\subset\mathbb{C}$ open set, there is a subharmonic function $\phi:U\to\mathbb{R}$ such that for any $c\in \mathbb{R}$, the set $\{z\in U: \phi(z)\le c\}$ is compact.
I don't know how to start, I only know some equivalent definitions for subharmonic functions and the problem seems too general because i have to find a $\phi$ for every $U.$
Hints: (i) If $f$ is holomorphic it's easy to show $|f|$ is subharmonic. So for example $|z|$ is subharmonic in the plane, and if $a\in\partial U$ then $\frac1{|z-a|}$ is subharmonic in $U$. (ii) If $u$ is usc and $u$ is the sup of a family of subharmonic functions it's easy to show that $u$ is subharmonic.
Indeed, say $u(z)=\sup_\alpha u_\alpha(z)$, where each $u_\alpha$ is subharmonic. If $z$ and $r$ are such that [...], then $$u_\alpha(z)\le\frac1{2\pi}\int_0^{2\pi}u_\alpha(z+re^{it})\,dt \le\frac1{2\pi}\int_0^{2\pi}u(z+re^{it})\,dt,$$hence $$u(z) \le\frac1{2\pi}\int_0^{2\pi}u(z+re^{it})\,dt.$$