= Solution
For a bounded domain, the <nondivergence-form elliptic operator> obeys the <weak maximum principle>: $Lu\geq0$ and $c\leq0$ imply
$$
\boxed{\sup_\Omega u\leq\max\{0,\sup_{\partial\Omega}u\}=\sup_{\partial\Omega}u^+.}
$$
Here $u^+=\max\{u,0\}$. Boundedness of $\Omega$ is needed for this formulation: it is not explicitly imposed in this question. On an unbounded domain one needs an appropriate condition at infinity instead. Without either condition the claim is false, as $u(x)=x_n$ and $L=\Delta$ on the half-space $\{x_n>0\}$ show.
For the bounded-domain proof choose $\gamma>0$ so large that $\lambda\gamma^2-\|b^1\|_\infty\gamma-\|c\|_\infty>0$. The positive exponential $h(x)=e^{\gamma x_1}$ then satisfies
$$
Lh=e^{\gamma x_1}\bigl(\gamma^2a^{11}+\gamma b^1+c\bigr)>0.
$$
Let $u_\varepsilon=u+\varepsilon h$, $M=\sup_{\partial\Omega}u^+$ and $H=\sup_{\overline\Omega}h<\infty$. If $u_\varepsilon$ had a value greater than $M+\varepsilon H$, it would attain a positive maximum at an interior point. There its <gradient> vanishes and its <Hessian matrix> is <negative semidefinite>. The symmetric part of $(a^{ij})$ is <positive-definite> by <uniform ellipticity>, so $a^{ij}D_{ij}u_\varepsilon\leq0$; also $cu_\varepsilon\leq0$. Hence $Lu_\varepsilon\leq0$, contradicting $Lu_\varepsilon=Lu+\varepsilon Lh>0$. Thus $u_\varepsilon\leq M+\varepsilon H$. Letting $\varepsilon\downarrow0$ proves the <weak maximum principle>, without requiring continuity of the coefficients.
For example, the unbounded-domain version follows by exhausting the domain with bounded truncations if $\limsup_{|x|\to\infty,\,x\in\Omega}u(x)\leq M$: the added spherical boundaries then have supremum at most $M+o(1)$.
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