I am currently trying to show $x\geq \sin(x)$ for $x\geq 0$. Therefore, I defined $f: \mathbb{R} \to \mathbb{R}, t\mapsto t-\sin(t)$. Then, $f'(t)\geq 0$ is true and thus $f$ is monotonically increasing. Furthermore, we reach equality, when $x=0$.
Is that enough reasoning? And how do I show that there is only one $x$ such that $x=\sin(x)$?
Consider the function $f(x) = \displaystyle \int_{0}^x (1-\cos(t))dt\ge 0, \forall x \ge 0\implies x -\sin x\ge 0\implies x \ge \sin x$ .
Yes, your argument is enough to show that $x \ge \sin(x)$ for all $x \ge 0$.
To show there is only one $x$ where equality holds: suppose there were some $y > 0$ with $f(y) = 0$. Since $f$ is monotone increasing, this can only happen if $f$ is identically zero on $[0,y]$. By symmetry $f$ is also identically zero on $[-y,0]$, thus on $[-y,y]$. But now note that $f'''(0) \ne 0$ which contradicts this.
Your proof is fine. For the second part: $f$ is strictly increasing, because the set $\{x|f'(x)=0\}$ contains no interval. For the proof of this claim, see this question, for example: Strictly increasing function and its derivative
