Proof that the minimum of a bounded differentiable real function occurs at a stationary point or at an endpoint I couldn't find a proof for this well-known result on this site...

The minimum of a bounded differentiable real function occurs at a stationary point or at an endpoint.

I prove the contrapositive to this statement in my answer below.
Alternative proofs are most welcome.
 A: If the domain is unbounded, say $(0,+\infty)$, then the question needs some clarification as the minimum of $x\to1/x$ is not attained.
If the domain is bounded, say $[0,1]$, the minimum exists by continuity of $f$ and a consequence of Bolzano-Weierstrass ("the image of a segment by a continuous function is a segment"). Let $c\in[0,1]$ be the point where the minimum is attained. If $c\notin\{0,1\}$, we look at
$$\frac{f(x)-f(c)}{x-c}$$
which is always non-negative if $x>c$ as $f(x)\ge f(c)$ holds. This proves that the limit $\lim_{x\to c^+}$ of the above quotient is non-negative. Similarly by looking at $x<c$, we see that $\lim_{x\to c^-}$ of the the above quotient is non-positive. Hence the limit is both non-positive and non-negative, i.e., $f'(c)=0$.
A: Let $f: [0,1] \to \mathbb{R}$ be a differentiable real function, and suppose that $c \in [0,1]$ is neither a stationary point nor an endpoint, but $c$ is a minimum. Then $f'(c)$ exists because $f$ is differentiable, but $f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}\neq 0,\ $ because $c$ is not a stationary point. Suppose WLOG that $f'(c)>0$. Then since $c$ is not the left endpoint, namely $0, $ and since $f'(c)>0, \ \exists k\ $ with $\ 0<k<c$ such that $f(k)<f(c)$, contradicting the minimality of $c$.
Admittedly I didn't prove the last line very formally, although I'm not sure how to, to be honest.
