 Solution

There is a missing hypothesis in the printed uniqueness claim: on an unbounded domain it is false without a condition at infinity. For example, on $\Omega =\{x_d>0\}$ take $f=0$ and $u_D=0$. Both $u=0$ and $u=x_d$ are classical solutions, although $f_u=0$.

For the intended bounded-domain statement, let $u,v\in C^2(\Omega )\cap C(\overline{\Omega })$ be solutions and put $w=u-v$. The derivative assumption makes $f$ nondecreasing in its first argument. On the open set $D=\{x\in \Omega :w(x)>0\}$,The difference $w$ vanishes on $\partial \Omega$, and continuity makes it zero on every interior boundary point of $D$ as well. Since $D$ is bounded, the weak maximum principle for subharmonic functions gives ${\sup }_{\overline{D}}w\le {\sup }_{\partial D}w=0$, contradicting $w>0$ there unless $D$ is empty. Thus $u\le v$. Interchanging $u,v$ gives the reverse inequality, proving $\boxed{u=v}$. An appropriate bound or decay at infinity can replace boundedness, but the printed hypotheses alone do not include it.