Consider the IVP given by
$ \frac{dx}{dt} = f(x), x(0)=0,$ where $f(x) =1$ if $x<0$ and $= -1 $ if $x \geq 0$
There exists no solution to this IVP on any interval $[0, T], T> 0$.
Incorrect proof: Suppose that there exists a solution to this IVP, given by $x: [0,T] \to \mathbb{R}, T >0$. Clearly $x(t) \neq 0$ for all $t \in (0, T]$ because $f$ is not continuous at 0. So suppose that $x(T) > 0.$ Then $x(t) > 0$ for all $t \in (0, T]$ and consequently $f(x(t)) = -1 $ for all $t \in (0, T]$. Then $ \frac{dx}{dt} = 1$ for all $t \in (0, T]$ and so $x(t) = -t$. So $x(t) = -t$ is a solution to the IVP but the proposition says that there is no solution, so we have a contradiction.
Where is the error in this proof?
In your proof attempt, you suppose that $x(t)>0$ for all $t>0$, and then use that assumption to construct a function such that $x(t)<0$ for all $t>0$, but that violates your original assumption!
Instead you might argue as follows: suppose $x(t)>0$ for all $t>0$. Then by definition of $f$ we have $\frac{dx}{dt}=-1$ for all $t>0$. Now since $x(0)=0$ and $x$ is continuous, we can find some $\delta>0$ such that $0<x(t)<\frac{T}{2}$ for all $0<t<\delta$. Then let $t<\min\{\delta,\frac{T}{2}\}$. Using the fundamental theorem of calculus, we get $x(t+\frac{T}{2}) = x(t) + \int_t^{t+\frac{T}{2}} \frac{dx}{dt}(y)dy = x(t) + \int_t^{t+\frac{T}{2}} (-1)dy = x(t) - \frac{T}{2} < 0$, but since $t+\frac{T}{2}<T$ by definition, this is a contradiction.
The case when you instead assume $x(t)<0$ for all $t>0$ is practically identical.