Following the wikipedia page of a Conservative force, I want to prove that given a force such that for any closed path $C$ we have $$ \oint_C \mathbf{F}\cdot \mathbf{r}=0 $$ there is a potential $U(\mathbf{r})$ such that $$ \mathbf{F}=-\nabla U(\mathbf{r}). $$
Proof. The webpage start by defining $$ U(\mathbf{r}):=-\int_{c}\mathbf{F}\cdot d \mathbf{r} $$ and conclude that the claim follows from the fundamental theorem of calculus.
I want to deepen the last step because the derivation is far from obvious for me.
First I adapt the definition of potential by choosing some zero potential point $\mathbf{x}_0$: $$ U(\mathbf{r}):=-\int_{\mathbf{x}_0}^\mathbf{x}\mathbf{F}\cdot d \mathbf{r} $$ then I introduce a parametrisation of the path $$ \mathbf{s}(u),\quad u\in[0,1] $$ Thus we get $$ -\nabla \left( -\int_{c}\mathbf{F}\cdot d \mathbf{r} \right) =\nabla\int_0^1 \left( F_x\frac{dx}{du} + F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du $$ For the $x$ component, intuitively but with a lot of errors, we must have something like this $$ \frac{d}{dx}\int_0^1 \left( F_x\frac{dx}{du} + F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du =\frac{du}{dx}\frac{d}{du}\int_0^1 \left( F_x\frac{dx}{du} + F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du\\ =\frac{du}{dx}\left( F_x\frac{dx}{du} + F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)_{\mathbf{x}_0}^\mathbf{x}=\cdots=F_x $$
How does the derivation looks like using formal mathematics, but remaining in calculus?
EDIT My solution going back to the definition of derivative:
Since no one else has written an answer here, I'll write a quick one now and I might edit it later if it isn't clear. I'll give an intuitive rather than a formal proof; it seems like this is what you're interested in.
Calculating the gradient
What does $-\nabla U(\textbf{r})$ mean for $U$ as defined in your question? Taking a derivative means that we are modifying the endpoint of the contour used in the integral and observing how $U$ changes when we do this.
The way we formalize this is to consider two contours, $c$ and $c + \delta c$, which end at slightly different locations $\textbf{r}$ and $\textbf{r} + \delta\textbf{r}$.
For simplicity, I'll assume that $\delta c$ is just a straight line from $\textbf{r}$ to $\textbf{r} + \delta\textbf{r}$. It actually doesn't matter because we are taking a derivative, so $\delta \textbf{r}$ is arbitrarily small. (Since $c$ is generally assumed to be piecewise differentiable, then if it is small it is basically a straight line.)
In a more formal treatment we would probably discuss the gradient by considering derivatives along an arbitrary direction, but for now just consider the derivative of $U$ with respect to $x$. In this case, $y$ and $z$ are not changing, so we can write write $\delta \textbf{r} = (\delta x, 0, 0)$. $$ \begin{align} \frac{\text{d}U}{\text{d}x} = - \frac{\text{d}}{\text{d}x} \int^\textbf{r} \textbf{F} \cdot \text{d}\textbf{r} &= - \lim_{\delta x \to 0} \frac{1}{\delta x} \left( \int^{\textbf{r} + \delta \textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} - \int^{\textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} \right) \\ &= - \lim_{\delta x \to 0} \frac{1}{\delta x} \int^{\textbf{r} + \delta \textbf{r}}_{\textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} \\ &= - \lim_{\delta x \to 0} \frac{1}{\delta x} \int_0^1 \left( F_x \delta x + F_y \cdot 0 + F_z \cdot 0 \right) \text{d}t \\ &= - \lim_{\delta x \to 0} \frac{1}{\delta x} F_x \delta x \\ &= - F_x \end{align} $$
The first line is just applying the definition of a derivative to this case. The second line is a basic property of integrals. The third line is just parameterizing, along with $\delta y = 0$ and $\delta z = 0$. From there it's trivial.
Now you could do the same thing for $y$ and $z$ and find that $$ \nabla U = - \textbf{F} $$ which is what you are trying to show. Basically, the thing does boil down to the fundamental theorem of calculus in one dimension, but you just have to set it up correctly.
Path-independence
Now, it's important to point out that there is one really important implicit assumption that I made above, which is that when I write the integral as $\int^\textbf{r}$, I am implicitly assuming that the result does not depend on the particular path taken. This is only true if $\textbf{F}$ is a "conservative" force, which is the whole point of your question; I'll briefly explain why this is justified with this assumption.
If we define a conservative force as one where every closed path has a contour integral of zero, as you have above, then it's very easy to show that this is true. Suppose that I have two paths $c_1$ and $c_2$ that start and end at the same point. Do they result in the same value of $U$ at the endpoint? Well, if you go "forwards" along $c_1$ and "backwards" along $c_2$, then you have a closed path and the integral along the total path $c_3$ has to be zero. This gives us $$ \int_{c_1} \textbf{F} \cdot \text{d}\textbf{r} - \int_{c_2} \textbf{F} \cdot \text{d}\textbf{r} = \int_{c_3} \textbf{F} \cdot \text{d}\textbf{r} = 0 $$ so the integrals along any two paths $c_1$ and $c_2$ with the same start and end points must be equal, and we can ignore the particular path taken, as I did in the derivation above.