Show that $\max_{\bar{U}} |u|\le C(\max_{\partial U} |g|+\max_{\bar{U}} |f|)$ 
Let $U$ be a bounded, open subset of $\mathbb{R}^n$. Prove that there exists a constant $C$, depending on only $U$, such that $$\max_{\bar{U}} |u|\le C(\max_{\partial U} |g|+\max_{\bar{U}} |f|)$$ wherever $u$ is a smooth solution of \begin{cases}-\Delta u=f &  \text{in }U\\ \quad \, \, \,  u=g & \text{on } \partial U \end{cases}

This is from PDE Evans (2nd edition), Chapter 2 Exercise 6. (Note that $\bar{U}$ is the closure of $U$.)
First, I tried proving for the boundary case, that is, suppose $x \in \partial U$. Then $u=g$. We have $$|u|=|g|\le \max_{\partial U} |g| \le \max_{\partial U}|g|+\max_{\bar{U}}|f|.$$ Hence, $|u|$ is bounded by $\max_{\partial U}|g|+\max_{\bar{U}}|f|$. So there exists a $C > 0$ such that $|u| \le C(\max_{\partial U}|g|+\max_{\bar{U}}|f|)$. Thus, $$\max_{\partial U} |u| \le C(\max_{\partial U}|g|+\max_{\bar{U}}|f|),$$ as required.  
Now, how can I prove for the interior case, that is, for $x \in U$? I wanted to use strong maximum principle, but that applies only to harmonic solutions, so I can't use that here.
 A: Here is a hint. You already noticed that we want to use Maximum principle and Maximum principle only happens on harmonic functions.
Let's divide your problem into too cases. Namely, 
\begin{cases}-\Delta u_1=0 &  \text{in }U\\ \quad \, \, \,  u_1=g & \text{on } \partial U \end{cases}
\begin{cases}-\Delta u_2=f &  \text{in }U\\ \quad \, \, \,  u_2=0 & \text{on } \partial U \end{cases}
You already have the estimation of $u_1$ by M-P. Now we work on $u_2$. 
Since $U$ is bounded, let's assume that $U\subset \{x\in R^N,\,\, |x\cdot \xi|\leq d\}$ where $|\xi|=1$ and $d>0$ is an constant.
Next we define 
$$ k(x):= e^{Lx\cdot \xi} $$
and we notice that 
$$ -\Delta k(x)=-L^2\xi_i^2e^{Lx\cdot\xi}=-L^2 e^{Lx\cdot\xi}\leq 1 $$
for large enough $L>0$.
Now we send
$$ v(x):= (e^{Ld}-e^{Lx\cdot \xi})\sup_{U}f^+ $$
Now we compute and find out
$$ -\Delta v(x)\geq \sup_{U}f^+ $$
and hence we have 
$$-\Delta (u_2(x)-v(x))\leq f-\sup_U f^+\leq 0 $$
hence we have $u-v$ is subharmonic and by M-P we have $u(x)\leq v(x)$ for all $x\in U$ and hence we have 
$$ \sup u_2\leq  C\sup f^+$$
Now write $u=u_1+u_2$, I am sure you could take it from here.
