Given a continuous map $f\colon [t_0,t_1]\times\mathbb{R}\to\mathbb{R}$, consider the initial value problem $$x'=f(t,x),\quad x(t_0)=x_0\in\mathbb{R}.$$ I have to show that $x\in C^0([t_0,t_1])$ is a solution of the IVP iff $x$ is a fixed-point of $$T\colon C^0([t_0,t_1])\to C^0([t_0,t_1]),\, (Tx)(t)=x_0+\int_{t_0}^t f(\tau,x(\tau))\operatorname{d}\tau.$$
How is this not immediately evident from the fundamental theorem of calculus? I have the feeling of missing something.
You are right: $x$ is a fixpoint of $T$ $ \iff$
$x(t)=x_0+\int_{t_0}^t f(\tau,x(\tau))\operatorname{d}\tau.$
And this is equivalent to $x'=f(t,x),\quad x(t_0)=x_0$
by the fundamental theorem of calculus.