Let $\Omega$ be a domain in $\mathbb{C}$ and let $h\in C^{\infty}\left(\Omega\right)$. Show that there exist $u\in C^{\infty}\left(\Omega\right)$ such that $u_{z\bar z}=h$, and if $h\left(\Omega\right)\subset \mathbb{R}$ then we can choose $u:\Omega\rightarrow\mathbb{R}$.
Using Runge's theorem I can prove that there exist $f\in C^{\infty}\left(\Omega\right)$ such that $\frac{\partial f}{\partial \bar z}=g$. (If $g$ has compact support it is trivial consequence of Cauchy-Green theorem, and if $g$ is unrestricted then we can construct sets as in proof of Mittag-Leffler theorem and use Runge's theorem). $(\star)$
How construct $u$ ? Is it a good idea to use $(\star)$ ?
Yes, it is a good idea to use $(\star)$ !
a) Using $(\star)$, choose $H\in C^{\infty}(\Omega)$ such that $H_{\bar z}=h$ and then using $(\star)$ again choose $U\in C^{\infty}(\Omega)$ such that $U_{\bar z}=\bar H$.
Then $\bar U_{z}= H$ and thus $\bar U_{z\bar z }= H_{\bar z}=h$ , so that for the required function you may take $u=\bar U$ which indeed satisfies $u_{z\bar z }= h$.
b) The above solution is of course completely valid if $h$ is real valued ($h(\Omega)\subset \mathbb R$), but if you insist on a real valued solution, just replace $u$ by $v=\frac {u+\bar u}{2}: \Omega \to \mathbb R$ which is $C^\infty$, real valued and still satisfies $v_{z\bar z }= h$.