Let $f:[0,1]\to\mathbb{R}$ be a real continuous function and for each $x\in[0,1)$ satisfies
$$ \liminf_{y \to x^+} \frac{f(y)-f(x)}{y-x} \leq 0 $$
We need to show that f is non-increasing on $[0,1]$. I attempted to prove by contradiction: Suppose otherwise, that $\exists \ x_1,x_2 \in [0,1]$ such that $x_1<x_2$, $f(x_1) < f(x_2)$. For any $y \in (f(x_1),f(x_2))$ (such a $y$ exists since $f$ is continous), let \begin{equation}x_0 = \sup\{x\in(x_1,x_2) : f(x) = y\} \end{equation} Then, by definition of $x_0$, $f(x) > y = f(x_0)$ for any $x\in(x_0,x_2)$, and hence \begin{equation}\liminf_{x \to x_0^+} \frac{f(x)-f(x_0)}{x-x_0} \geq 0 \end{equation}
I do not know how to proceed further, since the limit can be zero at $x_0$. I understand that I could always find another $y$ for which it is not zero, but i cannot seem to formalize it. Any help?
Given $0\leq x_0 < x_1 \leq 1$, we have to prove that $f(x_0) \geq f(x_1)$. Let us consider the auxiliary function $$ g(x) = f(x) - \frac{f(x_1) - f(x_0)}{x_1 - x_0} \, (x- x_0), \qquad x\in [x_0, x_1]. $$ Since $g$ is continuous in $[x_0, x_1]$, by Weierstrass' theorem there exists a minimum point $c\in [x_0, x_1]$ of $g$. Moreover, since $g(x_0) = g(x_1)$, we can assume without loss of generality that $c\in [x_0, x_1)$. Since $c$ is a minimum point of $g$, we clearly have that $$ \frac{g(y) - g(c)}{y - c}\geq 0, \qquad \forall y\in (c, x_1], $$ hence $$ 0\leq \liminf_{y\to c^+} \frac{g(y) - g(c)}{y - c} = \liminf_{y\to c^+} \frac{f(y) - f(c)}{y - c} - \frac{f(x_1) - f(x_0)}{x_1-x_0} \leq - \frac{f(x_1) - f(x_0)}{x_1-x_0}\,, $$ which in turn implies that $f(x_0) \geq f(x_1)$.