Take $\delta > 0$ to be fixed, and let's assume that $a>0$ (the proof for $a<0$ is analogous).
You can first prove that the end points of $[\bar{x}-\delta,\bar{x}+\delta]$ are mapped to the end points of $[\bar{x}-|a|\delta,\bar{x}+|a|\delta]$, just by applying $F$. Of course, this is not sufficient: it might be the case that some other point in $[\bar{x}-\delta,\bar{x}+\delta]$ ends up outside the target interval.
Therefore, the second thing should prove that any point $y \in (\bar{x},\bar{x}+\delta)$ ends up either 1) in $(\bar{x},\bar{x}+|a|\delta)$ or 2) in $(\bar{x}-|a|\delta,\bar{x})$.
Let's write $y$ as $y = \bar{x} + \epsilon$, with $0<\epsilon < \delta$. If we now apply the mapping $F$ to $y$, we get
\begin{equation}
F(y) = F(\bar{x}+\epsilon) = a (\bar{x} + \epsilon) + b = a \bar{x} + b + a \epsilon = F(\bar{x}) + a \epsilon = \bar{x} + a \epsilon,
\end{equation}
because $\bar{x}$ is the fixed point of $F$. We assumed $a>0$, so we see that
\begin{equation}
F(y) = \bar{x} + a \epsilon < \bar{x} + a \delta = \bar{x} + |a| \delta,
\end{equation}
and obviously $\bar{x} < F(y) = \bar{x} + a \epsilon$. So,
\begin{equation}
F(y) \in (\bar{x},\bar{x} + |a|\delta)
\end{equation}
for all $y \in (\bar{x},\bar{x} + \delta)$.
To conclude, we have proved that when $a>0$, the interval $[\bar{x},\bar{x}+\delta]$ is mapped by $F$ to the interval $[\bar{x},\bar{x} + |a| \delta]$. I'm sure you can now repeat the above for the other part of the interval, $[\bar{x} - \delta,\bar{x}]$. Also, you will have to consider the other possibility, namely that $a<0$. As I said, the proof in the $a<0$ case is analogous to that in the $a>0$ case.
