I have a function which is \begin{equation} F(x)= \begin{cases} f(x) & x \in [\underline{x},\bar{x})\\ \\ f(\bar{x}) & x=\bar{x} \end{cases} \end{equation}
The function $f(x)$ is strictly increasing in $[\underline{x},\bar{x})$, and $\lim_{x\to \bar{x}}f(x)=f(\bar{x})$. So I could conclude that $F(x)$ is continuous at $\bar{x}$.
Could I conclude that $f(\bar{x})>f(x)$ for all $x \in [\underline{x},\bar{x})$ ?
Yes. I will use $[a,b]$ instead of $[\underline{x},\overline{x}]$ in this demonstration.
Suppose there exist a $x \in [a,b[$ such that $f(x) \geq f(b)$
For every $\epsilon > 0$ such that $ \frac{b-x}{2} > \epsilon $, you have that
$$f(b-\epsilon) > f\left(\frac{b+x}{2}\right) > f(x) \geq f(b)$$
Now you make $\epsilon \to 0$, and you get
$$f(b) \geq f\left(\frac{b+x}{2}\right) > f(x) \geq f(b)$$
Contradiction