This answers @user3078439's question in his comment to my original answer.

First, a short argument to show that if $P,Q$ are domains and $f:P\to Q$ is Scott-continuous, then $f$ is order preserving.

Let $a\leq b \in P$. So $D:= \{a,b\}$ is a directed set with $b=\bigsqcup D$. Since $f$ is Scott-continuous, $f(D) = \{f(a), f(b)\}$ is also directed, so at least one of the relations $f(a) \leq f(b)$ or $f(b) \leq f(a)$ holds in $Q$. Now $f$ being Scott-continuous means $$f(b) = f(\bigsqcup D) = \bigsqcup f(D) = \sup\{f(a), f(b)\},$$ which directly implies $f(a)\leq f(b)$. So if $f:P\to Q$ is Scott-continuous, then it is order-preserving. (The converse of this statement does not hold.)

Since in your question the ordering on $Q:= \mathbb{IR}$ is given by $\supseteq$, I hope that I have answered your implicit question "if $f$ being Scott-continuous truly means that for all $x\leq y, f(x)\supseteq f(y)$" in the positive.

As to the question whether $f(-a)\supseteq f([-a,a])$ there is a negative answer: $f(-a)$ is just a real interval, namely $[f^-(-a), f^+(-a)]$. But $f([-a,a])$ is a **set of intervals**.

What is true, though, is that $f(-a)=[f^-(-a), f^+(-a)]$ is an interval that is a superset of every interval $f(x)=[f^-(x), f^+(x)]$ for all $x\in[-a,a]$. If that was the intent of your question in the comment, then there is a positive answer.