Show that $A_n= {\{x \in [a,b] : \exists y \in [a,b]\, |x−y| < \frac{1}{n} \ \text{and} \ F(y)−F(x) < p(y−x)}\}$ is open. In the proof of Lemma 7.17 from Bruckner's Real Analysis, it is claimed without a proof the following :

Let $F$ be continuous on $[a,b]$. For $n \in \mathbb{N}$ and $p \in \mathbb{R}$, let
$$A_n= {\{x \in [a,b] : \exists y \in [a,b]\, |x−y| < \frac{1}{n} \text{ and } F(y)−F(x) < p(y−x)}\}.$$
Since $F$ is continuous, each of the sets $A_n$ is open.

How $A_n$ is open?
 A: Given any $y \in [a, b]$, let $F_y : [a,b] \setminus  \{y\} \rightarrow \Bbb R $ be defined by $F_y(x) = \frac{F(y) - F(x)} { y-x}$. Note that $F_y$ is a continuous  function.
Now, let $x \in [a,b]$ and suppose $\exists y \in [a,b]\ \text{s.t.} |x−y| < \frac{1}{n} \ \text{and} \ F(y)−F(x) < p(y−x)$. Clearly $y \ne x$, otherwise
$$ 0= F(x)−F(x)<p(x-x)=0$$ Contradiction. So
\begin{align*} A_n&= \{x \in [a,b] : \exists y \in [a,b]\ \text{s.t.} |x−y| < \frac{1}{n} \ \text{and} \ F(y)−F(x) < p(y−x) \} = \\ 
& = \{x \in [a,b] : \exists y \in [a,b] \ \text{s.t.} |x−y| < \frac{1}{n} \ \text{and} \ F_y(x)< p \} = \\
&=  \{x \in [a,b] : \exists y \in [a,b] \ \text{s.t.} x \in B\left (y,\frac{1}{n} \right) \ \text{and} \ x\in  F_y^{-1}((-\infty, p ))\} = \\
&=\bigcup_{y \in [a,b]} \left (  B\left (y,\frac{1}{n} \right) \cap F_y^{-1}((-\infty, p ))  \right )
\end{align*}
Since $ B\left (y,\frac{1}{n} \right)$  is open in $[a,b]$ and $F_y^{-1}((-\infty, p ))$ is open in $[a,b] \setminus \{x\}$ which is open in $[a,b]$, we have that $ B\left (y,\frac{1}{n} \right)$  and $F_y^{-1}((-\infty, p ))$  are open in  $[a,b]$. So, $A_n$ is open in $[a,b]$.
