Consider the Cauchy problem $$\begin{cases}x'(t)=f(x) \\ x(0)=x_0\end{cases}$$ with $t\in \mathbb{R}$ and $f:I\subseteq\mathbb{R} \longrightarrow\mathbb{R}$.
If $f(x_0)=0$, then $x(t)=x_0$ is the solution of the equation. If $f(x_0)\neq 0$, then $f(x(t))\neq 0 \, \forall t$ in the maximal interval of definition of the solution, because the solution is unique.
I don't understand the bold part: supposing that $\exists t_1$ such that $f(x(t_1))=0$, what is the contradiction? I was thinking that, defining $x(t_1):=x_1$, if $x_1$ is a zero of $f$, then $x(t)=x_1$ can be another solution of the problem, which contradicts the unicity, but this is not true since $x(0)=x_1\neq x_0$, so $x(t)=x_1$ is not a solution of the problem. What am I missing?
If $x(t_1)=x_1$ and $f(x_1)=0$, then as you found also $\bar x(t)\equiv x_1$ is a solution.
Now you have the situation that both $x$ and $\bar x$ solve the Cauchy problem with initial condition $x(t_1)=x_1$. This is impossible. So you have to backtrack to the last assumption. Thus $x(t_1)=x_1$ or $\dot x(t_1)=f(x(t_1))=0$ is not possible. And if you want to force this, then $x(0)=x_0$ is not possible, but that was the problem to solve, so we can not throw out this condition.