Let $T>0$. Consider the following semilinear parabolic equation $$ \partial_tu(x,t)-\Delta u(x,t)=f(u(x,t)) \quad \forall (x,t) \in (0,1) \times (0,T), $$ with homogeneous Neumann boundary conditions and given initial conditions in $L^2(0,1)$.
Assume that $f:\mathbb{R}\longrightarrow\mathbb{R}$ is continuously differentiable. Prove that the solution to the stated PDE is unique.
Attempt:
Let $u_1,u_2$ be two solutions. Then, using Green formula, we obtain
$$
\begin{split}
\dfrac{1}{2} \dfrac{d}{dt}\int_{0}^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx&=\int_{0}^1 f(u_1(x,t))-f(u_2(x,t)) (u_1(x,t)-u_2(x,t))dx\\
&\leq C \int_{0}^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx
\end{split}
$$
where we have used the mean value theorem and such that $$C:=\underset{s \in [u_1(x,t),u_2(x,t)]}{\sup} f'(s)$$
which exists since $f$ is continuously differentiable.
By applying Gronwall inequality, we obtain $u_1=u_2.$
Is this proof correct? I can't spot the mistake.
$$ \dfrac{1}{2} \dfrac{d}{dt} \int_0^1 \vert u_1(x,t)-u_2(x,t)\vert^2 dx\leq \int_{0}^1C(x,t) \vert u_1(x,t)-u_2(x,t)\vert^2dx $$
The mistake is assuming that the constant $C$ is uniform.