Prove $\oint_\Gamma\vec\nabla f\cdot d\vec{r}=0$ when $\Gamma$ is the unit circle
Lucky you, the appearance of a totally wrong answer persuaded me to post a complete solution.
Fundamental Theorem of Calculus for line integrals: Suppose $\Omega\subset\Bbb R^2$ is open and $f\in C^1(\Omega)$. If $\gamma:[0,1]\to\Omega$ is a $C^1$ curve then $\int_\gamma\nabla f\cdot dr=f(\gamma(1))-f(\gamma(0))$.
(Hence if $\gamma$ is a closed curve (meaning $\gamma(1)=\gamma(0)$) the integral is $0$. With $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $\gamma(t)=(\cos(2\pi t),\sin(2\pi t))$ this proves the result you ask about; in particular what happens at the origin doesn't matter.)
Proof, even though it must be in the book: By definition, if $F$ is a vector field then $$\int_\gamma F\cdot dr=\int_0^1F(\gamma(t))\cdot\gamma'(t)\,dt.$$Now suppose that $F=\nabla f$, and set $$g(t)=f(\gamma(t)).$$The chain rule shows that $$F(\gamma(t))\cdot \gamma'(t)=g'(t),$$so $$\int_\gamma F\cdot dr=\int_0^1g'(t)\,dt=g(1)-g(0).$$
Corollary: If $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $F$ is the infamous vector field mentioned in the question there does not exist $f\in C^1(\Omega)$ with $F=\nabla f$.
(Indeed, in the language used in calculus books, $F$ is the standard example of a vector field which is closed but not exact.)